A GUIDE TO SQL
Eighth Edition
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A G U I D E TO S Q L
Eighth Edition
Philip J. Pratt Grand Valley State University Mary Z. Last University of Mary Hardin-Baylor
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A Guide to SQL, Eighth Edition Philip J. Pratt, Mary Z. Last Vice President, Publisher: Jack Calhoun Editor-in-Chief: Alex von Rosenberg Senior Acquisitions Editor: Charles McCormick, Jr.
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TABLE OF CONTENT S
Preface Chapter 1 Introduction to Premiere Products, Henry Books, and Alexamara Marina Group What Is a Database? The Premiere Products Database The Henry Books Database The Alexamara Marina Group Database Chapter Summary Key Terms Exercises
Chapter 2
Database Design Fundamentals
Database Concepts Relational Databases Entities, Attributes, and Relationships Functional Dependence Primary Keys Database Design Design Method Database Design Requirements Database Design Process Example Normalization First Normal Form Second Normal Form Third Normal Form Diagrams for Database Design Chapter Summary Key Terms Review Questions Exercises
Chapter 3
Creating Tables
Creating and Running SQL Commands Starting the Oracle Database Express Edition Entering Commands Creating a Table Correcting Errors in SQL Commands Dropping a Table Using Data Types Using Nulls Adding Rows to a Table The INSERT Command Inserting a Row that Contains Nulls Viewing Table Data Correcting Errors in a Table
viii 1 2 2 8 15 20 20 20
23 24 24 26 29 31 34 34 35 36 41 41 43 47 52 56 57 57 58
61 62 62 65 66 69 70 71 72 72 73 75 75 78
Saving SQL Commands Creating the Remaining Database Tables Describing a Table Chapter Summary Key Terms Review Questions Exercises
Chapter 4
Single-Table Queries
Constructing Simple Queries Retrieving Certain Columns and All Rows Retrieving All Columns and All Rows Using a WHERE Clause Using Compound Conditions Using the BETWEEN Operator Using Computed Columns Using the LIKE Operator Using the IN Operator Sorting Using the ORDER BY Clause Additional Sorting Options Using Functions Using the COUNT Function Using the SUM Function Using the AVG, MAX, and MIN Functions Using the DISTINCT Operator Nesting Queries Subqueries Grouping Using the GROUP BY Clause Using a HAVING Clause HAVING vs. WHERE Nulls Summary of SQL Clauses, Functions, and Operators Chapter Summary Key Terms Review Questions Exercises
Chapter 5
Multiple-Table Queries
Querying Multiple Tables Joining Two Tables Comparing joins, IN, and EXISTS Using the IN Operator Using the EXISTS Operator Using a Subquery Within a Subquery A Comprehensive Example Using an Alias Joining a Table to Itself Using a Self-Join on a Primary Key Column Joining Several Tables Set Operations ALL and ANY Special Operations Inner Join vi
80 82 87 89 89 89 90
97 98 98 100 100 103 106 107 110 111 112 112 113 114 114 115 116 117 120 121 123 123 125 125 128 128 130 130 131 132
135 135 136 139 140 141 142 144 146 147 148 150 152 157 160 160
Outer Join Product Chapter Summary Key Terms Review Questions Exercises
Chapter 6
Updating Data
Creating a New Table from an Existing Table Changing Existing Data in a Table Adding New Rows to an Existing Table Commit and Rollback Transactions Changing and Deleting Existing Rows Executing a Rollback Changing a Value in a Columnto Null Changing a Table’s Structure Making Complex Changes Dropping a Table Chapter Summary Key Terms Review Questions Exercises
Chapter 7
Database Administration
Creating and Using Views Using a View to Update Data Updating Row-and-Column Subset Views Updating Views Involving Joins Updating Views Involving Statistics Dropping a View Security Indexes Creating an Index Dropping an Index Creating Unique Indexes System Catalog Integrity Constraints in SQL Chapter Summary Key Terms Review Questions Exercises
Chapter 8
SQL Functions and Procedures
Using SQL in a Programming Environment Using Functions Character Functions Number Functions Working with Dates Concatenating Columns Stored Procedures Retrieving a Single Row and Column Error Handling Using Update Procedures
161 163 165 166 166 167
171 172 173 176 177 178 178 180 181 182 188 189 190 190 190 191
195 196 203 203 205 208 208 209 212 215 217 217 218 221 226 226 227 228
233 234 235 235 236 237 240 242 242 245 247 vii
Changing Data with a Procedure Deleting Data with a Procedure Selecting Multiple Rows with a Procedure Using a Cursor Opening a Cursor Fetching Rows from a Cursor Closing a Cursor Writing a Complete Procedure Using a Cursor Using More Complex Cursors Advantages of Cursors Using T-SQL in SQL Server Retrieving a Single Row and Column Changing Data with a Stored Procedure Deleting Data with a Stored Procedure Using a Cursor Using More Complex Cursors Using SQL in Microsoft Access Deleting Data with Visual Basic Running the Code Updating Data with Visual Basic Inserting Data with Visual Basic Finding Multiple Rows with Visual Basic Using a Trigger Chapter Summary Key Terms Review Questions Exercises
Appendix A
viii
247 248 249 249 250 251 252 252 254 255 256 256 256 257 257 258 259 259 260 261 262 262 264 267 268 268 269
SQL Reference
273
Aliases ALTER TABLE Column or Expression List (SELECT Clause) Computed Columns The DISTINCT Operator Functions COMMIT Conditions Simple Conditions Compound Conditions BETWEEN Conditions LIKE Conditions IN Conditions EXISTS Conditions ALL and ANY CREATE INDEX CREATE TABLE CREATE VIEW Data Types DELETE Rows DESCRIBE DROP INDEX DROP TABLE DROP VIEW GRANT INSERT INTO (Query)
273 273 274 274 274 275 275 275 275 276 276 276 276 277 277 277 278 279 279 280 281 281 281 282 282 282
INSERT INTO (Values) Integrity REVOKE ROLLBACK SELECT Subqueries UNION, INTERSECT, and MINUS UPDATE
Appendix B APPENDIX C Chapter Group Chapter Chapter Chapter Chapter Chapter Chapter Chapter
Index
How Do I Reference Answers to Odd-Numbered Review Questions
283 283 284 284 285 286 286 287
289 293
1—Introduction to Premiere Products, Henry Books, and Alexamara Marina 2—Database Design Fundamentals 3—Creating Tables 4—Single-Table Queries 5—Multiple-Table Queries 6—Updating Data 7—Database Administration 8—SQL Functions and Procedures
293 293 295 295 295 296 296 297
299
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P REFACE Structured Query Language (or SQL, which is pronounced “se-quel,” or “ess-cue-ell”) is a popular computer language that is used by diverse groups such as home computer users, owners of small businesses, end users in large organizations, and programmers. Although this text uses the SQL implementation in the Oracle Database 10g Express Edition as a vehicle for teaching SQL, its chapter material, examples, and exercises can be completed using any SQL implementation. A Guide to SQL, Eighth Edition is written for a wide range of teaching levels, from students taking introductory computer science classes to those students in advanced information systems classes. This text can be used for a standalone course on SQL or in conjunction with a database concepts text where students are required to learn SQL. The chapters in this text should be covered in order. Students should complete the end-ofchapter exercises and the examples within the chapters for maximum learning. Because the content of Chapter 8 assumes that the reader has had instruction or experience in at least one programming language, the instructor should determine whether students will understand its concepts. Students without a programming background will have difficulty understanding the topic of embedded SQL. Instructors can easily omit Chapter 8 from the text in situations where students are not comfortable with programming examples. The Eighth Edition builds on the success of previous editions by presenting basic SQL commands in the context of a business that uses SQL to manage orders, parts, customers, and sales reps. Like in previous editions, this edition uses Oracle as the vehicle to present SQL commands. Like the last edition, this edition addresses SQL in Access™ by showing the Access versions of the same commands when they differ from the Oracle versions. This new edition also shows SQL Server commands when they differ from the Oracle versions. Differences for Access and SQL Server users are featured in “User” notes, which make it easy for students to identify differences for the SQL implementation they are using. Students can download the Oracle 10g Database Express Edition from the Oracle Web site for free and use it to complete this text without having to purchase or install the full Oracle program. The Eighth Edition includes an entire chapter on database design, showing students how to create an appropriate design that satisfies a given set of requirements, and includes coverage of the important topics of stored procedures and triggers. The text also contains updated exercises for the Premiere Products, Henry Books, and Alexamara Marina Group cases.
DISTINGUISHING FEATURES Use of Examples Each chapter contains multiple examples that use SQL to solve a problem. Following each example, students will read about the commands that are used to solve the stated problem, and then they will see the SQL commands used to arrive at the solution. For most students, learning through examples is the most effective way to master material. For this reason, instructors should encourage students to read the chapters at the computer and input the commands shown in the figures.
Case Studies A running case study—Premiere Products—is presented in all of the examples within the chapters and in the first set of exercises at the end of each chapter. Although the database is small in order to be manageable, the examples and exercises for the Premiere Products database simulate what a real business can accomplish using SQL commands. Using the same case study as examples within the chapter and in the end-of-chapter exercises ensures a high level of continuity to reinforce learning. A second case study—the Henry Books database—is used in a second set of exercises at the end of each chapter. A third case study—the Alexamara Marina Group database—is used in a third set of exercises at the end of each chapter. The second and third case studies give students a chance to venture out “on their own” without the direct guidance of examples from the text.
Question and Answer Sections A special type of exercise, called a Q&A, is used throughout the book. These exercises force students to consider special issues and understand important questions before continuing with their study. The answer to each Q&A appears after the question. Students are encouraged to formulate their own answers before reading the ones provided in the text to ensure that they understand new material before proceeding.
“User” Notes for Access™ and SQL Server Users When an SQL command has a different use or format in Access or SQL Server, it appears in a User note. When you encounter a User note for the SQL implementation you are using, be sure to read its contents. You might also review the User notes for other SQL implementations so you are aware of the differences that occur from one implementation of SQL to another.
Review Material A Summary and Key Terms list appear at the end of each chapter, followed by Review Questions that test students’ recall of the important points in the chapter and occasionally test their ability to apply what they have learned. The answers to the odd-numbered Review Questions are provided in Appendix C. Each chapter also contains exercises related to the Premiere Products, Henry Books, and Alexamara Marina Group databases.
Appendices Three appendices appear at the end of this text. Appendix A is an SQL reference that describes the purpose and syntax for the major SQL commands featured in the text. Students can use Appendix A to identify how and when to use important commands quickly. The SQL reference appendix contains references to specific pages in the text where the command is discussed to make it easy for students to find additional information when they need to refer back to the section in the book where the topic is covered. Appendix B includes a “How Do I” reference, which lets students cross-reference the appropriate section in Appendix A by searching for the answer to a question. Appendix C includes answers to the odd-numbered Review Questions.
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Relationship to Concepts of Database Management, Sixth Edition For database courses featuring SQL, this SQL text can be bundled with Concepts of Database Management, Sixth Edition by Pratt and Adamski (Course Technology). The data and pedagogy between the two texts is consistent, and the instructor’s manuals for both books include feedback and suggestions for using the texts together.
Instructor Support The Eighth Edition includes a package of proven supplements for instructors and students. The Instructor’s Resources offer a detailed electronic Instructor’s Manual, figure files, Microsoft® PowerPoint® presentations, and the ExamView® Test Bank. The Instructor’s Manual includes suggestions and strategies for using this text, as well as answers to Review Questions and solutions to the end-of-chapter exercises. Figure files allow instructors to create their own presentations using figures appearing in the text. Instructors can also take advantage of lecture presentations provided on PowerPoint slides; these presentations follow each chapter’s coverage precisely, include chapter figures, and can be customized. ExamView is a powerful objective-based test generator that enables instructors to create paper, LAN, or Web-based tests from test banks designed specifically for this Course Technology text. Users can utilize the ultra-efficient QuickTest Wizard to create tests in less than five minutes by taking advantage of Course Technology’s question banks, or can customize their own exams from scratch. The Instructor’s Resources include copies of the databases for the Premiere Products, Henry Books, and Alexamara Marina Group cases in Microsoft Access 2007 and 2003 formats and script files to create the tables and data in these databases in Oracle and SQL Server. These files are provided so instructors have the choice of assigning exercises in which students create the databases used in this text and load them with data, or they can provide the starting Access databases or Oracle or SQL Server script files to students to automate and simplify these tasks.
ORGANIZATION OF THE TEXT The text contains eight chapters and three appendices, which are described in the following sections.
Chapter 1: Introduction to Premiere Products, Henry Books, and Alexamara Marina Group Chapter 1 introduces the three database cases that are used throughout the text: Premiere Products, Henry Books, and Alexamara Marina Group. Many Q&A exercises are provided throughout the chapter to ensure that students understand how to manipulate the database on paper before they begin working in SQL.
Chapter 2: Database Design Fundamentals Chapter 2 covers important concepts and terminology associated with relational databases, functional dependence, and primary keys, followed by a method for designing a database to satisfy a given set of requirements. It also illustrates the normalization process for finding and correcting a variety of potential problems in database designs. Finally, it shows how to represent database designs graphically using entity-relationship diagrams. Preface
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Chapter 3: Creating Tables In Chapter 3, students begin using a DBMS by creating and running SQL commands to create tables, use data types, and add rows to tables. Chapter 3 also discusses the role of and use of nulls.
Chapter 4: Single-Table Queries Chapter 4 is the first of two chapters on using SQL commands to query a database. The queries in Chapter 4 all involve single tables. Included in this chapter are discussions of simple and compound conditions; computed columns; the SQL BETWEEN, LIKE, and IN operators; using SQL aggregate functions; nesting queries; grouping data; and retrieving columns with null values.
Chapter 5: Multiple-Table Queries Chapter 5 completes the discussion of querying a database by demonstrating queries that join more than one table. Included in this chapter are discussions of the SQL IN and EXISTS operators, nested subqueries, using aliases, joining a table to itself, SQL set operations, and the use of the ALL and ANY operators. The chapter also includes coverage of various types of joins.
Chapter 6: Updating Data In Chapter 6, students learn how to use the SQL COMMIT, ROLLBACK, UPDATE, INSERT, and DELETE commands to update table data. Students also learn how to create a new table from an existing table and how to change the structure of a table. The chapter also includes coverage of transactions, including both their purpose and implementation.
Chapter 7: Database Administration Chapter 7 covers the database administration features of SQL, including the use of views; granting and revoking database privileges to users; creating, dropping, and using an index; using and obtaining information from the system catalog; and using integrity constraints to control data entry.
Chapter 8: SQL Functions and Procedures Chapter 8 begins with a discussion of some important SQL functions that act on single rows. Students will also learn how to use PL/SQL and T-SQL to cover the process of embedding SQL commands in another language. Included in this chapter are discussions of using embedded SQL to insert new rows and change and delete existing rows. Also included is a discussion of how to retrieve single rows using embedded SQL commands and how to use cursors to retrieve multiple rows. Chapter 8 also includes a section showing some techniques for using SQL in Visual Basic (Access). The chapter concludes with a discussion of triggers.
Appendix A: SQL Reference Appendix A includes a command reference for all the major SQL clauses and operators that are featured in the chapters. Students can use Appendix A as a quick resource when constructing commands. Each command includes a short description, a table that shows xiv
the required and optional clauses and operators, and an example and its results. It also contains a reference to the pages in the text where the command is covered.
Appendix B: How Do I Reference Appendix B provides students with an opportunity to ask a question, such as “How do I delete rows?”, and to identify the appropriate section in Appendix A to use to find the answer. Appendix B is extremely valuable when students know what task they want to accomplish but can’t remember the exact SQL command they need.
Appendix C: Answers to Odd-Numbered Review Questions Answers to the odd-numbered Review Questions in each chapter appear in this appendix so students can make sure that they are completing the Review Questions correctly.
GENERAL NOTES TO THE STUDENT You can download the databases used in this text from www.cengage.com. The Access data files for this book include three Access 2003 databases (Premiere Products.mdb, Henry Books.mdb, and Alexamara Marina Group.mdb), which you can open in Access 2000, 2002, or 2003; and three Access 2007 databases (Premiere Products.accdb, Henry Books.accdb, and Alexamara Marina Group.accdb), which you can open in Access 2007. The data files also include script files for Oracle and SQL Server that you can use to create or drop the Premiere Products, Henry Books, and Alexamara Marina Group databases. The script files saved in the Oracle folder have the following functions: Oracle-Alexamara.sql: Creates all the tables in the Alexamara Marina Group database and adds all the data. Run this script file to create the Alexamara Marina Group database. (Note: This script file assumes you have not previously created any of the tables in the database. If you have created any of the tables, you should run the Oracle-DropAlexamara.sql script prior to running the Oracle-Alexamara.sql script.) Oracle-Henry.sql: Creates all the tables in the Henry Books database and adds all the data. Run this script file to create the Henry Books database. (Note: This script file assumes you have not previously created any of the tables in the database. If you have created any of the tables, you should run the Oracle-DropHenry.sql script prior to running the Oracle-Henry.sql script.) Oracle-Premiere.sql: Creates all the tables in the Premiere Products database and adds all the data. Run this script file to create the Premiere Products database. (Note: This script file assumes you have not previously created any of the tables in the database. If you have created any of the tables, you should run the Oracle-DropPremiere.sql script prior to running the Oracle-Premiere.sql script.) Oracle-DropAlexamara.sql: Drops (deletes) all the tables and data in the Alexamara Marina Group database. Oracle-DropHenry.sql: Drops (deletes) all the tables and data in the Henry Books database.
Preface
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Oracle-DropPremiere.sql: Drops (deletes) all the tables and data in the Premiere Products database. The script files saved in the SQL Server folder have the following functions: SQLServer-Alexamara.sql: Creates all the tables in the Alexamara Marina Group database and adds all the data. Run this script file to create the Alexamara Marina Group database. (Note: This script file assumes you have not previously created any of the tables in the database. If you have created any of the tables, you should run the SQLServer-DropAlexamara.sql script prior to running the SQLServer-Alexamara.sql script.) SQLServer-Henry.sql: Creates all the tables in the Henry Books database and adds all the data. Run this script file to create the Henry Books database. (Note: This script file assumes you have not previously created any of the tables in the database. If you have created any of the tables, you should run the SQLServer-DropHenry.sql script prior to running the SQLServer-Henry.sql script.) SQLServer-Premiere.sql: Creates all the tables in the Premiere Products database and adds all the data. Run this script file to create the Premiere Products database. (Note: This script file assumes you have not previously created any of the tables in the database. If you have created any of the tables, you should run the SQLServer-DropPremiere.sql script prior to running the SQLServer-Premiere.sql script.) SQLServer-DropAlexamara.sql: Drops (deletes) all the tables and data in the Alexamara Marina Group database. SQLServer-DropHenry.sql: Drops (deletes) all the tables and data in the Henry Books database. SQLServer-DropPremiere.sql: Drops (deletes) all the tables and data in the Premiere Products database. For details on running script files in Oracle or SQL Server, check with your instructor. You can also refer to Chapter 3 in the text for information about creating and using scripts. For information about downloading the Oracle Database 10g Express Edition software, please visit the Oracle Web site. For information about SQL Server 2005, please visit the Microsoft Web site. Information about Microsoft Access is also available at the Microsoft Web site.
Embedded Questions In many places, you’ll find Q&A sections to ensure that you understand some crucial material before you proceed. In some cases, the questions are designed to give you the chance to consider some special concept in advance of its actual presentation. In all cases, the answer to each question appears immediately after the question. You can simply read the question and its answer, but you will benefit from taking time to determine the answer to the question before checking your answer against the one given in the text.
End-of-Chapter Material The end-of-chapter material consists of a Summary, a Key Terms list, Review Questions, and exercises for the Premiere Products, Henry Books, and Alexamara Marina Group databases. The Summary briefly describes the material covered in the chapter. The Review Questions require you to recall and apply the important material in the chapter. The answers to the odd-numbered Review Questions appear in Appendix C so you can check
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your progress. The Premiere Products, Henry Books, and Alexamara Marina Group exercises test your knowledge of the chapter material; your instructor will assign one or more of these exercises for you to complete.
ACKNOWLEDGMENTS We would like to acknowledge several individuals for their contributions in the preparation of this text. We appreciate the efforts of the following individuals who reviewed the manuscript and made many helpful suggestions: Vickee Stedham, St. Petersburg College; Bill Kloepfer, Golden Gate University; Georgia Brown, Northern Illinois University; Gary Savard, Champlain College; Stephen Cerovski, Coleman College; Ricardo Herrera, Vanier College and Concordia University; Eugenia Fernandez, Indiana University-Purdue University Indianapolis; Danny Yakimchuk, University College of Cape Breton; Paul Leidig, Grand Valley State University; Misty Vermaat, Purdue University Calumet; Lorna Bowen St. George, Old Dominion University; and George Federman, Santa Barbara Community College. The efforts of the following members of the staff at Course Technology have been invaluable and have made this text possible: Charles McCormick, Senior Acquisitions Editor; Kate Hennessy, Product Manager; Matt Hutchinson, Content Project Manager; Marisa Taylor, Project Manager; and GreenPen Quality Assurance testers. We have once again had the great pleasure to work with an absolutely amazing Developmental Editor, Jessica Evans, on several books. Thanks for all your efforts, Jess. You’re the best! We’ve said it before, but it is just as true as ever!
Preface
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CHAPTER
1
I N T RO D UCT IO N TO P REMIERE PRO D UC TS, H E N R Y B OOKS, AN D A L EX AMAR A MAR IN A G ROU P LEARNING
OBJECTIVES
Objectives ●
Introduce Premiere Products, a company whose database is used as the basis for many of the examples throughout the text
●
Introduce Henry Books, a company whose database is used as a case that runs throughout the text
●
Introduce Alexamara Marina Group, a company whose database is used as an additional case that runs throughout the text
INTRODUCTION In this chapter, you will examine the database requirements of Premiere Products, a company that will be used in the examples throughout the text. Then you will examine the database requirements for Henry Books and Alexamara Marina Group, whose databases are featured in the exercises that appear at the end of each chapter.
2
WHAT IS A DATABASE? Throughout this text, you will work with databases for three organizations: Premiere Products, Henry Books, and Alexamara Marina Group. A database is a structure that contains different categories of information and the relationships between these categories. The Premiere Products database, for example, contains information about categories such as sales representatives (sales reps), customers, orders, and parts. The Henry Books database contains information about categories such as books, publishers, authors, and branches. The Alexamara Marina Group database contains information about categories such as marinas, slips and the boats in them, service categories, and service requests. Each database also contains relationships between categories. For example, the Premiere Products database contains information that relates sales reps to the customers they represent and customers to the orders they have placed. The Henry Books database contains information that relates publishers to the books they publish and authors to the books they have written. The Alexamara Marina Group database contains information that relates the boats in the slips at the marina to the owners of the boats. As you work through the chapters in this text, you will learn more about these databases and how to view and update the information they contain. As you read each chapter, you will see examples from the Premiere Products database. At the end of each chapter, your instructor might assign the exercises for the Premiere Products, Henry Books, or Alexamara Marina Group databases.
THE PREMIERE PRODUCTS DATABASE The management of Premiere Products, a distributor of appliances, housewares, and sporting goods, has determined that the company’s recent growth no longer makes it feasible to maintain customer, order, and inventory data using its manual systems. With the data stored in a database, management will be able to ensure that the data is current and more accurate than in the present manual systems. In addition, managers will be able to obtain answers to their questions concerning the data in the database easily and quickly, with the option of producing a variety of useful reports. Management has determined that Premiere Products must maintain the following information about its sales reps, customers, and parts inventory in the new database: ●
●
●
The number, last name, first name, address, total commission, and commission rate for each sales rep The customer number, name, address, current balance, and credit limit for each customer, as well as the number of the sales rep who represents the customer The part number, description, number of units on hand, item class, number of the warehouse where the item is stored, and unit price for each part in inventory
Premiere Products also must store information about orders. Figure 1-1 shows a sample order.
Chapter 1
3
PREMIERE PRODUCTS
ORDER: 2 1 6 1 7
Heading
DATE: 1 0 / 2 3 / 2 0 1 0
SALES REP: 6 5
CUSTOMER: 6 0 8
J u a n Pe r e z
J o h n s o n ’s D e p a r t m e n t S t o r e 372 Oxford Sheldon FL 33553 PART NUMBER
Order lines
BV06 CD52
PART DESCRIPTION
Home Gym Microwave Oven
NUMBER ORDERED
PRICE
2 4
794.95 150.00
TOTAL
1589.90 600.00
Body
Extensions
Footing
FIGURE 1-1
ORDER TOTAL >>
2189.90
Sample order
The sample order shown in Figure 1-1 has three sections: ●
●
●
The heading (top) of the order contains the company name; the order number and date; the customer’s number, name, and address; and the sales rep’s number and name. The body of the order contains one or more order lines, sometimes called line items. Each order line contains a part number, a part description, the number of units of the part ordered, and the quoted price for the part. Each order line also contains a total, usually called an extension, which is the result of multiplying the number ordered by the quoted price. Finally, the footing (bottom) of the order contains the order total.
Premiere Products also must store the following items in the database for each customer’s order: ●
●
●
For each order, the database must store the order number, the date the order was placed, and the number of the customer that placed the order. The customer’s name and address and the number of the sales rep who represents the customer are stored with the customer information. The name of the sales rep is stored with the sales rep information. For each order, the database must store the order number, the part number, the number of units ordered, and the quoted price for each order line. The part description is stored with the information about parts. The result of multiplying the number of units ordered by the quoted price is not stored because the database can calculate it when needed. The overall order total is not stored. Instead, the database calculates the total whenever an order is printed or displayed on the screen.
Figure 1-2 shows sample data for Premiere Products.
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
4
REP REP_ LAST_ FIRST_ NUM NAME NAME
STREET
CITY
STATE
ZIP
624 Randall
COMMISSION
RATE
20
Kaiser
Valerie
Grove
FL
33321
$20,542.50
0.05
35
Hull
Richard 532 Jackson
Sheldon
FL
33553
$39,216.00
0.07
65
Perez
Juan
Fillmore
FL
33336
$23,487.00
0.05
1626 Taylor
CUSTOMER CUSTOMER_ CUSTOMER_ NUM NAME
CITY
STATE ZIP
BALANCE
$6,550.00
CREDIT_ REP_ LIMIT NUM
148
Al’s Appliance 2837 and Sport Greenway
Fillmore
FL
33336
282
Brookings Direct
3827 Devon
Grove
FL
33321
356
Ferguson’s
382 Wildwood Northfield FL
33146
$5,785.00
$7,500.00 65
408
The Everything Shop
1828 Raven
Crystal
FL
33503
$5,285.25
$5,000.00 35
462
Bargains Galore
3829 Central
Grove
FL
33321
$3,412.00 $10,000.00 65
$7,500.00 20
$431.50 $10,000.00 35
524
Kline’s
838 Ridgeland Fillmore
FL
33336 $12,762.00 $15,000.00 20
608
Johnson’s Department Store
372 Oxford
FL
33553
$2,106.00 $10,000.00 65
687
Lee’s Sport 282 Evergreen Altonville FL and Appliance
32543
$2,851.00
$5,000.00 35
725
Deerfield’s Four Seasons
282 Columbia Sheldon
FL
33553
$248.00
$7,500.00 35
842
All Season
28 Lakeview
FL
33321
$8,221.00
$7,500.00 20
FIGURE 1-2
Chapter 1
STREET
Sheldon
Grove
Sample data for Premiere Products
ORDERS
ORDER_LINE
ORDER_ ORDER_ NUM DATE
21608
10/20/2010 148
21610
10/20/2010 356
21613
10/21/2010 408
21614
10/21/2010 282
21617
10/23/2010 608
21619
10/23/2010 148
21623
10/23/2010 608
5
ORDER_NUM PART_NUM NUM_ORDERED QUOTED_PRICE
CUSTOMER_ NUM
21608
AT94
11
$21.95
21610
DR93
1
$495.00
21610
DW11
1
$399.99
21613
KL62
4
$329.95
21614
KT03
2
$595.00
21617
BV06
2
$794.95
21617
CD52
4
$150.00
21619
DR93
1
$495.00
21623
KV29
2
$1,290.00
PART PART_NUM DESCRIPTION
ON_HAND CLASS WAREHOUSE
PRICE
AT94
Iron
50 HW
3
$24.95
BV06
Home Gym
45 SG
2
$794.95
CD52
Microwave Oven
32 AP
1
$165.00
DL71
Cordless Drill
21 HW
3
$129.95
DR93
Gas Range
8 AP
2
$495.00
DW11
Washer
12 AP
3
$399.99
FD21
Stand Mixer
22 HW
3
$159.95
KL62
Dryer
12 AP
1
$349.95
KT03
Dishwasher
8 AP
3
$595.00
KV29
Treadmill
9 SG
2
$1,390.00
FIGURE 1-2
Sample data for Premiere Products (continued)
In the REP table, you see that there are three sales reps, whose numbers are 20, 35, and 65. The name of sales rep 20 is Valerie Kaiser. Her street address is 624 Randall. She lives in Grove, Florida, and her zip code is 33321. Her total commission is $20,542.50, and her commission rate is five percent (0.05). In the CUSTOMER table, 10 Premiere Products customers are identified with the numbers 148, 282, 356, 408, 462, 524, 608, 687, 725, and 842. The name of customer number 148 is Al’s Appliance and Sport. This customer’s address is 2837 Greenway in Fillmore, Florida, with a zip code of 33336. The customer’s current balance is $6,550.00, and its credit limit is $7,500.00. The number 20 in the REP_NUM column indicates that Al’s Appliance and Sport is represented by sales rep 20 (Valerie Kaiser). Skipping to the table named PART, you see that there are 10 parts, whose part numbers are AT94, BV06, CD52, DL71, DR93, DW11, FD21, KL62, KT03, and KV29. Part AT94 is an iron, and the company has 50 units of this part on hand. Irons are in item class HW (housewares) and are stored in warehouse 3. The price of an iron is $24.95. Other item classes are AP (appliances) and SG (sporting goods).
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
6
Moving back to the table named ORDERS, you see that there are seven orders, which are identified with the numbers 21608, 21610, 21613, 21614, 21617, 21619, and 21623. Order number 21608 was placed on October 20, 2010, by customer 148 (Al’s Appliance and Sport).
NOTE In some database systems, the word order has a special purpose. Having a table named ORDER could cause problems in such systems. For this reason, Premiere Products uses the table name ORDERS instead of ORDER.
The table named ORDER_LINE might seem strange at first glance. Why do you need a separate table for the order lines? Could they be included in the ORDERS table? The answer is technically yes. You could structure the table named ORDERS as shown in Figure 1-3. Notice that this table contains the same orders as shown in Figure 1-2, with the same dates and customer numbers. In addition, each table row in Figure 1-3 contains all the order lines for a given order. Examining the fifth row, for example, you see that order 21617 has two order lines. One of these order lines is for two BV06 parts at $794.95 each, and the other order line is for four CD52 parts at $150.00 each. ORDERS ORDER_NUM ORDER_DATE CUSTOMER_NUM PART_NUM NUM_ORDERED QUOTED_PRICE
21608
10/20/2010
148
AT94
11
$21.95
21610
10/20/2010
356
DR93 DW11
1 1
$495.00 $399.99
21613
10/21/2010
408
KL62
4
$329.95
21614
10/21/2010
282
KT03
2
$595.00
21617
10/23/2010
608
BV06 CD52
2 4
$794.95 $150.00
21619
10/23/2010
148
DR93
1
$495.00
21623
10/23/2010
608
KV29
2
$1,290.00
FIGURE 1-3
Alternative ORDERS table structure
Q&A Question: How is the information from Figure 1-2 represented in Figure 1-3? Answer: Examine the ORDER_LINE table shown in Figure 1-2 and note the sixth and seventh rows. The sixth row indicates that there is an order line on order 21617 for two BV06 parts at $794.95 each. The seventh row indicates that there is an order line on order 21617 for four CD52 parts at $150.00 each. Thus, the information that you find in Figure 1-3 is represented in Figure 1-2 in two separate rows rather than in one row.
It might seem inefficient to use two rows to store information that could be represented in one row. There is a problem, however, with the arrangement shown in Figure Chapter 1
1-3—the table is more complicated. In Figure 1-2, there is a single entry at each location in the table. In Figure 1-3, some of the individual positions within the table contain multiple entries, making it difficult to track the information between columns. In the row for order number 21617, for example, it is crucial to know that the BV06 corresponds to the 2 in the NUM_ORDERED column (not the 4) and that it corresponds to the $794.95 in the QUOTED_PRICE column (not the $150.00). In addition, a more complex table raises practical issues, such as: ● ●
●
7
How much room do you allow for these multiple entries? What happens when an order has more order lines than you have allowed room for? For a given part, how do you determine which orders contain order lines for that part?
Although none of these problems is unsolvable, they do add a level of complexity that is not present in the arrangement shown in Figure 1-2. In Figure 1-2, there are no multiple entries to worry about, it does not matter how many order lines exist for any order, and finding every order that contains an order line for a given part is easy (just look for all order lines with the given part number in the PART_NUM column). In general, this simpler structure is preferable, and that is why order lines appear in a separate table. To test your understanding of the Premiere Products data, use Figure 1-2 to answer the following questions.
Q&A Question: What are the numbers of the customers represented by Valerie Kaiser? Answer: 148, 524, and 842. (Look up the REP_NUM value of Valerie Kaiser in the REP table and obtain the number 20. Then find all customers in the CUSTOMER table that have the number 20 in the REP_NUM column.)
Q&A Question: What is the name of the customer that placed order 21610, and what is the name of the rep who represents this customer? Answer: Ferguson’s is the customer; Juan Perez is the sales rep. (Look up the CUSTOMER_NUM value in the ORDERS table for order number 21610 and obtain the number 356. Then find the customer in the CUSTOMER table with the CUSTOMER_NUM value of 356. Using the REP_NUM value, which is 65, find the name of the rep in the REP table.)
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
8
Q&A Question: List all parts that appear in order 21610. For each part, give the description, number ordered, and quoted price. Answer: Part number: DR93; part description: Gas Range; number ordered: 1; and quoted price: $495.00. Also, part number: DW11; part description: Washer; number ordered: 1; and quoted price: $399.99. (Look up each ORDER_LINE table row in which the order number is 21610. Each of these rows contains a part number, the number ordered, and the quoted price. Use the part number to look up the corresponding part description in the PART table.)
Q&A Question: Why is the QUOTED_PRICE column part of the ORDER_LINE table? Can’t you just use the part number and look up the price in the PART table? Answer: If the QUOTED_PRICE column did not appear in the ORDER_LINE table, you would need to obtain the price for a part on an order line by looking up the price in the PART table. Although this approach is reasonable, it prevents Premiere Products from charging different prices to different customers for the same part. Because Premiere Products wants the flexibility to quote and charge different prices to different customers, the QUOTED_PRICE column is included in the ORDER_LINE table. If you examine the ORDER_LINE table, you will see cases in which the quoted price matches the actual price in the PART table and cases in which it differs. For example, in order number 21608, Al’s Appliance and Sport bought 11 irons, and Premiere Products charged only $21.95 per iron, rather than the regular price of $24.95.
THE HENRY BOOKS DATABASE Ray Henry is the owner of a bookstore chain named Henry Books. Like the management of Premiere Products, Ray has decided to store his data in a database. He wants to achieve the same benefits; that is, he wants to ensure that his data is current and accurate. In addition, he wants to be able to ask questions concerning the data and to obtain answers to these questions easily and quickly. In running his chain of bookstores, Ray gathers and organizes information about branches, publishers, authors, and books. Figure 1-4 shows sample branch and publisher data for Henry Books. Each branch has a number that uniquely identifies the branch. In addition, Ray tracks the branch’s name, location, and number of employees. Each publisher has a code that uniquely identifies the publisher. In addition, Ray tracks the publisher’s name and city.
Chapter 1
BRANCH
9
BRANCH_NUM BRANCH_NAME
1 Henry Downtown
BRANCH_LOCATION NUM_EMPLOYEES
16 Riverview
10
2 Henry On The Hill 1289 Bedford
6
3 Henry Brentwood
Brentwood Mall
15
4 Henry Eastshore
Eastshore Mall
9
PUBLISHER PUBLISHER_CODE PUBLISHER_NAME
CITY
AH
Arkham House
Sauk City WI
AP
Arcade Publishing
New York
BA
Basic Books
Boulder CO
BP
Berkley Publishing
Boston
BY
Back Bay Books
New York
CT
Course Technology
Boston
FA
Fawcett Books
New York
FS
Farrar Straus and Giroux
New York
HC
HarperCollins Publishers
New York
JP
Jove Publications
New York
JT
Jeremy P. Tarcher
Los Angeles
LB
Lb Books
New York
MP
McPherson and Co.
Kingston
PE
Penguin USA
New York
PL
Plume
New York
PU
Putnam Publishing Group
New York
RH
Random House
New York
SB
Schoken Books
New York
SC
Scribner
New York
SS
Simon and Schuster
New York
ST
Scholastic Trade
New York
TA
Taunton Press
Newtown CT
TB
Tor Books
New York
TH
Thames and Hudson
New York
TO
Touchstone Books
Westport CT
VB
Vintage Books
New York
WN
W.W. Norton
New York
WP
Westview Press
Boulder CO
FIGURE 1-4
Sample branch and publisher data for Henry Books
Figure 1-5 shows sample author data for Henry Books. Each author has a number that uniquely identifies the author. In addition, Ray records each author’s last and first names. Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
10
AUTHOR AUTHOR_NUM AUTHOR_LAST AUTHOR_FIRST
1 Morrison
Toni
2 Solotaroff
Paul
3 Vintage
Vernor
4 Francis
Dick
5 Straub
Peter
6 King
Stephen
7 Pratt
Philip
8 Chase
Truddi
9 Collins
Bradley
10 Heller
Joseph
11 Wills
Gary
12 Hofstadter
Douglas R.
13 Lee
Harper
14 Ambrose
Stephen E.
15 Rowling
J.K.
16 Salinger
J.D.
17 Heaney
Seamus
18 Camus
Albert
19 Collins, Jr.
Bradley
20 Steinbeck
John
21 Castelman
Riva
22 Owen
Barbara
23 O’Rourke
Randy
24 Kidder
Tracy
25 Schleining
Lon
FIGURE 1-5
Sample author data for Henry Books
Figure 1-6 shows sample book data for Henry Books. Each book has a code that uniquely identifies the book. For each book, Ray also tracks the title, publisher, book type, price, and whether the book is a paperback.
Chapter 1
BOOK
11
BOOK_CODE TITLE
PUBLISHER_ TYPE PRICE PAPERBACK CODE
0180
A Deepness in the Sky
TB
SFI
$7.19 Y
0189
Magic Terror
FA
HOR
$7.99 Y
0200
The Stranger
VB
FIC
$8.00 Y
0378
Venice
SS
ART
$24.50 N
079X
Second Wind
PU
MYS
$24.95 N
0808
The Edge
JP
MYS
$6.99 Y
1351
Dreamcatcher: A Nvel
SC
HOR
$19.60 N
1382
Treasure Chests
TA
ART
$24.46 N
138X
Beloved
PL
FIC
$12.95 Y
2226
Harry Potter and the Prisoner of Azkaban
ST
SFI
$13.96 N
2281
Van Gogh and Gauguin
WP
ART
$21.00 N
2766
Of Mice and Men
PE
FIC
$6.95 Y
2908
Electric Light
FS
POE
$14.00 N
3350
Group: Six People in Search of a BP Life
PSY
$10.40 Y
3743
Nine Stories
LB
FIC
$5.99 Y
3906
The Soul of a New Machine
BY
SCI
$11.16 Y
5163
Travels with Charley
PE
TRA
$7.95 Y
5790
Catch-22
SC
FIC
$12.00 Y
6128
Jazz
PL
FIC
$12.95 Y
6328
Band of Brothers
TO
HIS
$9.60 Y
669X
A Guide to SQL
CT
CMP
$37.95 Y
6908
Franny and Zooey
LB
FIC
$5.99 Y
7405
East of Eden
PE
FIC
$12.95 Y
7443
Harry Potter and the Goblet of Fire
ST
SFI
$18.16 N
7559
The Fall
VB
FIC
$8.00 Y
8092
Godel, Escher, Bach
BA
PHI
$14.00 Y
8720
When Rabbit Howls
JP
PSY
$6.29 Y
9611
Black House
RH
HOR
$18.81 N
9627
Song of Solomon
PL
FIC
$14.00 Y
9701
The Grapes of Wrath
PE
FIC
$13.00 Y
9882
Slay Ride
JP
MYS
$6.99 Y
9883
The Catcher in the Rye
LB
FIC
$5.99 Y
9931
To Kill a Mockingbird
HC
FIC
$18.00 N
FIGURE 1-6
Sample book data for Henry Books
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
12
To check your understanding of the relationship between publishers and books, answer the following questions.
Q&A Question: Who published Jazz? Which books did Jove Publications publish? Answer: Plume published Jazz. In the row in the BOOK table for Jazz (see Figure 1-6), find the publisher code PL. Examining the PUBLISHER table (see Figure 1-4), you see that PL is the code assigned to Plume. Jove Publications published The Edge, When Rabbit Howls, and Slay Ride. To find the books published by Jove Publications, find its code (JP) in the PUBLISHER table. Next, find all records in the BOOK table for which the publisher code is JP.
The table named WROTE, as shown in Figure 1-7, relates books to the authors who wrote them. The SEQUENCE column indicates the order in which the authors of a particular book are listed on the cover. The table named INVENTORY in the same figure is used to indicate the number of copies of a particular book that are currently on hand at a particular branch of Henry Books. The first row, for example, indicates that there are two copies of the book with the code 0180 at branch 1. INVENTORY
WROTE BOOK_CODE AUTHOR_NUM SEQUENCE
BOOK_CODE BRANCH_NUM ON_HAND
0180
1
0180
1
2
0189
5
1
0189
2
2
0200
18
1
0200
1
1
0378
11
1
0200
2
3
079X
4
1
0378
3
2
0808
4
1
079X
2
1
1351
6
1
079X
3
2
1382
23
2
079X
4
3
1382
25
1
0808
2
1
138X
1
1
1351
2
4
2226
15
1
1351
3
2 1
2281
9
2
1382
2
2281
19
1
138X
2
3
2766
20
1
2226
1
3
2908
17
1
2226
3
2
3350
2
1
2226
4
1
3743
16
1
2281
4
3
FIGURE 1-7
Chapter 1
3
Sample data that relates books to authors and books to branches for Henry Books
WROTE
INVENTORY
13
BOOK_CODE AUTHOR_NUM SEQUENCE
BOOK_CODE BRANCH_NUM ON_HAND
3906
24
1
2766
3
2
5163
20
1
2908
1
3
5790
10
1
2908
4
1
6128
1
1
3350
1
2
6328
14
1
3743
2
1
669X
7
1
3906
2
1
6908
16
1
3906
3
2
7405
20
1
5163
1
1
7443
15
1
5790
4
2
7559
18
1
6128
2
4
8092
12
1
6128
3
3
8720
8
1
6328
2
2
9611
5
2
669X
1
1
9611
6
1
6908
2
2
9627
1
1
7405
3
2
9701
20
1
7443
4
1
9882
4
1
7559
2
2
9883
16
1
8092
3
1
9931
13
1
8720
1
3
FIGURE 1-7
9611
1
2
9627
3
5
9627
4
2
9701
1
2
9701
2
1
9701
3
3
9701
4
2
9882
3
3
9883
2
3
9883
4
2
9931
1
2
Sample data that relates books to authors and books to branches for Henry Books (continued)
To check your understanding of the relationship between authors and books, answer the following questions.
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
14
Q&A Question: Who wrote Black House? (Make sure to list the authors in the correct order.) Which books did Toni Morrison write? Answer: Stephen King and Peter Straub wrote Black House. First examine the BOOK table (see Figure 1-6) to find the book code for Black House (9611). Next, look for all rows in the WROTE table in which the book code is 9611. There are two such rows. In one row, the author number is 5, and in the other, it is 6. Then, look in the AUTHOR table to find the authors who have been assigned the numbers 5 and 6. The answers are Peter Straub (5) and Stephen King (6). The sequence number for author number 5 is 2, and the sequence number for author number 6 is 1. Thus, listing the authors in the proper order results in Stephen King and Peter Straub. Toni Morrison wrote Beloved, Jazz, and Song of Solomon. To find the books written by Toni Morrison, look up her author number (1) in the AUTHOR table. Then look for all rows in the WROTE table for which the author number is 1. There are three such rows. The corresponding book codes are 138X, 6128, and 9627. Looking up these codes in the BOOK table, you find that Toni Morrison wrote Beloved, Jazz, and Song of Solomon.
Q&A Question: A customer in branch 1 wants to purchase The Soul of a New Machine. Is this book currently in stock at branch 1? Answer: No. Looking up the code for The Soul of a New Machine in the BOOK table, you find it is 3906. To find out how many copies are in stock at branch 1, look for a row in the INVENTORY table with 3906 in the BOOK_CODE column and 1 in the BRANCH_NUM column. Because there is no such row, branch 1 doesn’t have any copies of The Soul of a New Machine.
Q&A Question: You would like to obtain a copy of The Soul of a New Machine for this customer. Which other branches currently have this book in stock, and how many copies does each branch have? Answer: Branch 2 has one copy, and branch 3 has two copies. You already know that the code for The Soul of a New Machine is 3906. (If you did not know the book code, you would look it up in the BOOK table.) To find out which branches currently have copies, look for rows in the INVENTORY table with 3906 in the BOOK_CODE column. There are two such rows. The first row indicates that branch 2 currently has one copy. The second row indicates that branch 3 currently has two copies.
Chapter 1
THE ALEXAMARA MARINA GROUP DATABASE
15
Alexamara Marina Group offers in-water boat storage to owners by providing boat slips that owners can rent on an annual basis. Alexamara owns two marinas: Alexamara East and Alexamara Central. Each marina has several boat slips available. Alexamara also provides a variety of boat repair and maintenance services to the boat owners who rent the slips. Alexamara stores the data it needs to manage its operations in a relational database containing the tables described in the following section. Alexamara stores information about its two marinas in the MARINA table shown in Figure 1-8. A marina number uniquely identifies each marina. The table also includes the marina name, street address, city, state, and zip code. MARINA MARINA_NUM NAME
ADDRESS
1
Alexamara East
108 2nd Ave. Brinman
FL
32273
2
Alexamara Central 283 Branston W. Brinman FL
32274
FIGURE 1-8
CITY
STATE ZIP
Sample marina data for Alexamara Marina Group
Alexamara stores information about the boat owners to whom it rents slips in the OWNER table shown in Figure 1-9. An owner number that consists of two uppercase letters followed by a two-digit number uniquely identifies each owner. For each owner, the table also includes the last name, first name, address, city, state, and zip code. OWNER OWNER_ LAST_ NAME NUM
FIRST_NAME
AD57
Adney
Bruce and Jean 208 Citrus
AN75
Anderson Bill
18 Wilcox
BL72
Blake
Mary
2672 Commodore Bowton
FL
31313
EL25
Elend
Sandy and Bill
462 Riverside
Rivard
FL
31062
FE82
Feenstra
Daniel
7822 Coventry
Kaleva
FL
32521
JU92
Juarez
Maria
8922 Oak
Rivard
FL
31062
KE22
Kelly
Alyssa
5271 Waters
Bowton
FL
31313
NO27
Norton
Peter
2811 Lakewood
Lewiston
FL
32765
SM72
Smeltz
Becky and Dave 922 Garland
Glander Bay FL
31044
TR72
Trent
Ashton
Bay Shores
30992
FIGURE 1-9
ADDRESS
922 Crest
CITY
STATE
ZIP
Bowton
FL
31313
Glander Bay FL
31044
FL
Sample owner data for Alexamara Marina Group
Each marina contains slips that are identified by slip numbers. Marina 1 (Alexamara East) has two sections (A and B) and slips are numbered within each section. Thus, slip Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
16
numbers at marina 1 consist of the letter A or B followed by a number (for example, A3 or B2). At marina 2 (Alexamara Central), a number (1, 2, 3) identifies each slip. Information about the slips in the marinas is contained in the MARINA_SLIP table shown in Figure 1-10. Each row in the table contains a slip ID that identifies the particular slip. The table also contains the marina number and slip number, the length of the slip (in feet), the annual rental fee, the name of the boat currently occupying the slip, the type of boat, and the boat owner’s number. MARINA_SLIP SLIP_ID MARINA_ SLIP_ LENGTH RENTAL_ BOAT_NAME BOAT_TYPE OWNER_ NUM NUM FEE NUM
1 1
A1
40 $3,800.00 Anderson II
Sprite 4000
AN75
2 1
A2
40 $3,800.00 Our Toy
Ray 4025
EL25
3 1
A3
40 $3,600.00 Escape
Sprite 4000
KE22
4 1
B1
30 $2,400.00 Gypsy
Dolphin 28
JU92
5 1
B2
30 $2,600.00 Anderson III Sprite 3000
AN75
6 2
1
25 $1,800.00 Bravo
Dolphin 25
AD57
7 2
2
25 $1,800.00 Chinook
Dolphin 22
FE82
8 2
3
25 $2,000.00 Listy
Dolphin 25
SM72
9 2
4
30 $2,500.00 Mermaid
Dolphin 28
BL72
10 2
5
40 $4,200.00 Axxon II
Dolphin 40
NO27
11 2
6
40 $4,200.00 Karvel
Ray 4025
TR72
FIGURE 1-10
Sample data about slips at Alexamara Marina Group
Alexamara provides boat maintenance service for owners at its two marinas. The types of service provided are stored in the SERVICE_CATEGORY table shown in Figure 1-11. A category number uniquely identifies each service that Alexamara performs. The table also contains a description of the category. SERVICE_CATEGORY CATEGORY_NUM CATEGORY_DESCRIPTION
1 Routine engine maintenance 2 Engine repair 3 Air conditioning 4 Electrical systems 5 Fiberglass repair 6 Canvas installation 7 Canvas repair 8 Electronic systems (radar, GPS, autopilots, etc.)
FIGURE 1-11
Chapter 1
Sample data about service categories at Alexamara Marina Group
Information about the services requested by owners is stored in the SERVICE_REQUEST table shown in Figure 1-12. Each row in the table contains a service ID that identifies each service request. The slip ID identifies the location (marina number and slip number) of the boat to be serviced. For example, the slip ID on the second row is 5. As indicated in the MARINA_SLIP table in Figure 1-10, the slip ID 5 identifies the boat in marina 1 and slip number B2. The SERVICE_REQUEST table also contains the category number of the service to be performed, plus a description of the specific service to be performed, and a description of the current status of the service. It also contains the estimated number of hours required to complete the service. For completed jobs, the table contains the actual number of hours it took to complete the service. If another appointment is required to complete additional service, the appointment date appears in the NEXT_SERVICE_DATE column.
17
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
18
SERVICE_REQUEST SERVICE_ SLIP_ CATEGORY_ DESCRIPTION ID ID NUM
EST_ SPENT_ NEXT_ HOURS HOURS SERVICE_ DATE
1
1
3 Air conditioner periodically stops with code indicating low coolant level. Diagnose and repair.
Technician has verified the problem. Air conditioning specialist has been called.
4
2
7/12/2010
2
5
4 Fuse on port motor blown on two occasions. Diagnose and repair.
Open
2
0
7/12/2010
3
4
1 Oil change and general routine maintenance (check fliud levels, clean sea strainers, etc.).
Service call has been scheduled.
1
0
7/16/2010
4
1
2
2
0
7/13/2010
5
3
5 Open pockets at base of two stantions.
Technician has completed the initial filling of the open pockets. Will complete the job after the initial fill has had sufficient time to dry.
4
2
7/13/2010
6
11
4 Electric-flush system periodically stops functioning. Diagnose and repair.
Open
3
0
7
6
2 Engine overheating. Loss of coolant. Diagnose and repair.
Open
2
0
7/13/2010
8
6
2 Heat exchanger not operating correctly.
Technician has determined that the exchanger is faulty. New exchanger has been ordered.
4
1
7/17/2010
9
7
6 Canvas severely damaged in windstorm. Order and install new canvas.
Open
8
0
7/16/2010
10
2
8 Install new GPS and chart plotter.
Scheduled
7
0
7/17/2010
FIGURE 1-12 Chapter 1
STATUS
Engine oil level Open has been dropping drastically. Diagnose and repair.
Sample data about service requests at Alexamara Marina Group
SERVICE_REQUEST
19
SERVICE_ SLIP_ CATEGORY_ DESCRIPTION ID ID NUM
STATUS
EST_ SPENT_ NEXT_ HOURS HOURS SERVICE_ DATE
11
2
3 Air conditioning unit shuts down with HHH showing on the control panel.
Technician not able to replicate the problem. Air conditioning unit ran fine through multiple tests. Owner to notify technician if the problem recurs.
1
1
12
4
8 Both speed and depth readings on data unit are significantly less than the owner thinks they should be.
Technician has scheduled appointment with owner to attempt to verify the problem.
2
0
7/16/2010
13
8
2 Customer describes engine as making a clattering sound.
Technician suspects problem with either propeller or shaft and has scheduled the boat to be pulled from the water for further investigation.
5
2
7/12/2010
14
7
5 Owner accident caused damage to forward portion of port side.
Technician has scheduled repair.
6
0
7/13/2010
15
11
7 Canvas leaks around zippers in heavy rain. Install overlap around zippers to prevent leaks.
Overlap has been created. Installation has been scheduled.
8
3
7/17/2010
FIGURE 1-12
Sample data about service requests at Alexamara Marina Group (continued)
The Alexamara Marina Group exercises at the end of this chapter will give you a chance to check your understanding of the data in this database.
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
20
Chapter Summary ●
Premiere Products is an organization whose information requirements include sales reps, customers, parts, orders, and order lines.
●
Henry Books is an organization whose information requirements include branches, publishers, authors, books, inventory, and author sequences.
●
Alexamara Marina Group is an organization whose information requirements include marinas, owners, slips, service categories, and service requests.
Key Terms database
Exercises Premiere Products Answer each of the following questions using the Premiere Products data shown in Figure 1-2. No computer work is required. 1.
List the names of all customers that have a credit limit of $7,500 or less.
2.
List the order numbers for orders placed by customer number 608 on 10/23/2010.
3.
List the part number, part description, and on-hand value for each part in item class SG. (Hint: On-hand value is the result of multiplying the number of units on hand by the price.)
4.
List the part number and part description of all parts that are in item class HW.
5.
How many customers have a balance that exceeds their credit limit?
6.
What is the part number, description, and price of the least expensive part in the database?
7.
For each order, list the order number, order date, customer number, and customer name.
8.
For each order placed on October 21, 2010, list the order number, customer number, and customer name.
9.
List the sales rep number and name for every sales rep who represents at least one customer with a credit limit of $10,000.
10.
For each order placed on October 21, 2010, list the order number, part number, part description, and item class for each part ordered.
Henry Books Answer each of the following questions using the Henry Books data shown in Figures 1-4 through 1-7. No computer work is required.
Chapter 1
1.
List the name of each publisher that is located in New York.
2.
List the name of each branch that has at least nine employees.
3.
List the book code and title of each book that has the type FIC.
4.
List the book code and title of each book that has the type FIC and that is in paperback.
5.
List the book code and title of each book that has the type FIC or whose publisher code is SC.
6.
List the book code and title of each book that has the type MYS and a price of less than $20.
7.
Customers who are part of a special program get a 10 percent discount off regular book prices. For the first five books in the BOOK table, list the book code, title, and discounted price. (Use the PRICE column to calculate the discounted price.)
8.
Find the name of each publisher containing the word and.
9.
List the book code and title of each book that has the type FIC, MYS, or ART.
10.
How many books have the type SFI?
11.
Calculate the average price for books that have the type ART.
12.
For each book published by Penguin USA, list the book code and title.
13.
List the book code, book title, and units on hand for each book in branch number 3.
21
Alexamara Marina Group Answer each of the following questions using the Alexamara Marina Group data shown in Figures 1-8 through 1-12. No computer work is required. 1.
List the owner number, last name, and first name of every boat owner.
2.
List the last name and first name of every owner located in Bowton.
3.
List the marina number and slip number for every slip whose length is equal to or less than 30 feet.
4.
List the marina number and slip number for every boat with the type Dolphin 28.
5.
List the slip number for every boat with the type Dolphin 28 that is located in marina 1.
6.
List the boat name for each boat located in a slip whose length is between 25 and 30 feet.
7.
List the slip number for every slip in marina 1 whose annual rental fee is less than $3,000.
8.
Labor is billed at the rate of $60 per hour. List the slip ID, category number, estimated hours, and estimated labor cost for every service request. To obtain the estimated labor cost, multiply the estimated hours by 60. Use the column name ESTIMATED_COST for the estimated labor cost.
9.
List the marina number and slip number for all slips containing a boat with the type Sprite 4000, Sprite 3000, or Ray 4025.
10.
How many Dolphin 25 boats are stored at both marinas?
11.
For every boat, list the marina number, slip number, boat name, owner number, owner’s first name, and owner’s last name.
12.
For every service request for routine engine maintenance, list the slip ID, the description, and the status.
13.
For every service request for routine engine maintenance, list the slip ID, marina number, slip number, estimated hours, spent hours, owner number, and owner’s last name.
Introduction to Premiere Products, Henry Books, and Alexamara Marina Group
This page intentionally left blank
CHAPTER
2
D A TABASE D ESIGN F U N D AMEN TALS LEARNING
OBJECTIVES
Objectives ●
Understand the terms entity, attribute, and relationship
●
Understand the terms relation and relational database
●
Understand functional dependence and identify when one column is functionally dependent on another
●
Understand the term primary key and identify primary keys in tables
●
Design a database to satisfy a set of requirements
●
Convert an unnormalized relation to first normal form
●
Convert tables from first normal form to second normal form
●
Convert tables from second normal form to third normal form
●
Create an entity-relationship diagram to represent the design of a database
INTRODUCTION In Chapter 1, you reviewed the tables and columns in the Premiere Products, Henry Books, and Alexamara Marina Group databases that you will use to complete the rest of this text. The process of determining the particular tables and columns that will comprise a database is known as database design. In this chapter, you will learn a method for designing a database to satisfy a set of requirements. In the process, you will learn how to identify the tables and columns in the database. You also will learn how to identify the relationships between the tables.
This chapter begins by examining some important concepts related to databases. It also presents 24
the design method using the set of requirements that Premiere Products identified to produce the appropriate database design. The chapter then examines the process of normalization, in which you identify and fix potential problems in database designs. Finally, you will learn a way of visually representing the design of a database.
DATA BA S E C O N C E P T S Before learning how to design a database, you need to be familiar with some important database concepts related to relational databases, which are the types of databases you examined in Chapter 1 and that you will use throughout the rest of this text. The terms entity, attribute, and relationship are important to understand when designing a database; the concepts of functional dependence and primary keys are critical when learning about the database design process.
Relational Databases A relational database is a collection of tables like the ones you examined for Premiere Products in Chapter 1 and that also appear in Figure 2-1. Formally, these tables are called relations, and this is how this type of database gets its name.
Chapter 2
REP REP_ LAST_ FIRST_ NUM NAME NAME
STREET
CITY
STATE ZIP
COMMISSION
RATE
20
Kaiser
Valerie
624 Randall
Grove
FL
33321
$20,542.50
0.05
35
Hull
Richard 532 Jackson
Sheldon
65
Perez
Juan
Fillmore
FL
33553
$39,216.00
0.07
FL
33336
$23,487.00
0.05
25
1626 Taylor
CUSTOMER CUSTOMER_ CUSTOMER_ NUM NAME
STREET
CITY
STATE ZIP
BALANCE
$6,550.00
148
Al’s Appliance 2837 and Sport Greenway
Fillmore
FL
33336
282
Brookings Direct
3827 Devon
Grove
FL
33321
CREDIT_ REP_ LIMIT NUM
$7,500.00 20
$431.50 $10,000.00 35
356
Ferguson’s
382 Wildwood Northfield FL
33146
$5,785.00
$7,500.00 65
408
The Everything Shop
1828 Raven
Crystal
FL
33503
$5,285.25
$5,000.00 35
462
Bargains Galore
3829 Central
Grove
FL
33321
$3,412.00 $10,000.00 65
524
Kline’s
838 Ridgeland Fillmore
FL
33336 $12,762.00 $15,000.00 20
608
Johnson’s Department Store
372 Oxford
FL
33553
$2,106.00 $10,000.00 65
687
Lee’s Sport 282 Evergreen Altonville FL and Appliance
32543
$2,851.00
$5,000.00 35
725
Deerfield’s Four Seasons
282 Columbia Sheldon
FL
33553
$248.00
$7,500.00 35
842
All Season
28 Lakeview
FL
33321
$8,221.00
$7,500.00 20
Grove
ORDER_LINE
ORDERS ORDER_ ORDER_ NUM DATE
21608
Sheldon
ORDER_NUM PART_NUM NUM_ORDERED QUOTED_PRICE
CUSTOMER_ NUM
10/20/2010 148
21610
10/20/2010 356
21613
10/21/2010 408
21614
10/21/2010 282
21617
10/23/2010 608
21619
10/23/2010 148
21623
10/23/2010 608
21608
AT94
11
$21.95
21610
DR93
1
$495.00
21610
DW11
1
$399.99
21613
KL62
4
$329.95
21614
KT03
2
$595.00
21617
BV06
2
$794.95
21617
CD52
4
$150.00
21619
DR93
1
$495.00
21623
KV29
2
$1,290.00
PART PART_NUM DESCRIPTION
ON_HAND CLASS WAREHOUSE
PRICE
AT94
Iron
50 HW
3
$24.95
BV06
Home Gym
45 SG
2
$794.95
CD52
Microwave Oven
32 AP
1
$165.00
DL71
Cordless Drill
21 HW
3
$129.95
DR93
Gas Range
8 AP
2
$495.00
DW11
Washer
12 AP
3
$399.99
FD21
Stand Mixer
22 HW
3
$159.95
KL62
Dryer
12 AP
1
$349.95
KT03
Dishwasher
8 AP
3
$595.00
KV29
Treadmill
9 SG
2
$1,390.00
FIGURE 2-1 Sample data for Premiere Products Database Design Fundamentals
NOTE 26
The names of columns and tables in this text follow a common naming convention in which column names use uppercase letters and replace spaces between words with underscores (_). For example, Premiere Products uses the column named LAST_NAME to store last names and the column named CREDIT_LIMIT to store credit limits.
Entities, Attributes, and Relationships There are some terms and concepts that are very important for you to know when working in the database environment. The terms entity, attribute, and relationship are fundamental when discussing databases. An entity is like a noun; it is a person, place, thing, or event. The entities of interest to Premiere Products, for example, are such things as customers, orders, and sales reps. The entities that are of interest to a school include students, faculty, and classes; a real estate agency is interested in clients, houses, and agents; and a used car dealer is interested in vehicles, customers, and manufacturers. An attribute is a property of an entity. The term is used here exactly as it is used in everyday English. For the entity person, for example, the list of attributes might include such things as eye color and height. For Premiere Products, the attributes of interest for the entity customer are such things as name, address, city, and so on. For the entity faculty at a school, the attributes would be such things as faculty number, name, office number, phone, and so on. For the entity vehicle at a car dealership, the attributes are such things as the vehicle identification number, model, color, year, and so on. A relationship is the association between entities. There is an association between customers and sales reps, for example, at Premiere Products. A sales rep is associated with all of his or her customers, and a customer is associated with his or her sales rep. Technically, you say that a sales rep is related to all of his or her customers, and a customer is related to his or her sales rep. The relationship between sales reps and customers is an example of a one-to-many relationship because one sales rep is associated with many customers, but each customer is associated with only one sales rep. (In this type of relationship, the word many is used in a way that is different from everyday English; it might not always mean a large number. In this context, for example, the term many means that a sales rep might be associated with any number of customers. That is, one sales rep can be associated with zero, one, or more customers.) How does a relational database handle entities, attributes of entities, and relationships between entities? Entities and attributes are fairly simple. Each entity has its own table. In the Premiere Products database, there is one table for sales reps, one table for customers, and so on. The attributes of an entity become the columns in the table. In the table for sales reps, for example, there is a column for the sales rep number, a column for the sales rep’s first name, and so on. What about relationships? At Premiere Products, there is a one-to-many relationship between sales reps and customers (each sales rep is related to the many customers that he or she represents, and each customer is related to the one sales rep who represents the customer). How is this relationship implemented in a relational database? Consider Figure 2-1 again. If you want to determine the name of the sales rep who represents Brookings Direct (customer number 282), you would locate the row for Brookings Direct Chapter 2
in the CUSTOMER table and determine that the value for REP_NUM is 35. Then you would look for the row in the REP table on which the REP_NUM is 35. The one rep with REP_NUM 35 is Richard Hull, who represents Brookings Direct. On the other hand, if you want to determine the names of all the customers of the rep named Valerie Kaiser, you would locate the row for Valerie Kaiser in the REP table and determine that the value in the REP_NUM column is 20. Then you would look for all the rows in the CUSTOMER table on which the REP_NUM is 20. After identifying Valerie Kaiser’s rep number, you find that the many customers she represents are numbered 148 (Al’s Appliance and Sport), 524 (Kline’s), and 842 (All Season). You implement these relationships by having common columns in two or more tables. The REP_NUM column in the REP table and the REP_NUM column in the CUSTOMER table are used to implement the relationship between sales reps and customers. Given a sales rep, you can use these columns to determine all the customers that he or she represents; given a customer, you can use these columns to find the sales rep who represents the customer. In this context, a relation is essentially a two-dimensional table. If you consider the tables shown in Figure 2-1, however, you can see that certain restrictions are placed on relations. Each column has a unique name, and entries within each column should “match” this column name. For example, if the column name is CREDIT_LIMIT, all entries in that column must be credit limits. Also, each row should be unique—when two rows are identical, the second row does not provide any new information. For maximum flexibility, the order of the columns and rows should be immaterial. Finally, the table’s design should be as simple as possible by restricting each position to a single entry and by preventing multiple entries (also called repeating groups) in an individual location in the table. Figure 2-2 shows a table design that includes repeating groups.
27
ORDERS ORDER_ ORDER_ DATE NUM
CUSTOMER_ NUM
PART_ NUM
NUM_ QUOTED_ ORDERED PRICE
21608
10/20/2010 148
AT94
11
$21.95
21610
10/20/2010 356
DR93 DW11
1 1
$495.00 $399.99
21613
10/21/2010 408
KL62
4
$329.95
21614
10/21/2010 282
KT03
2
$595.00
21617
10/23/2010 608
BV06 CD52
2 4
$12.95 $150.00
21619
10/23/2010 148
DR93
1
$495.00
21623
10/23/2010 608
KV29
2
$325.99
FIGURE 2-2
Table with repeating groups
Figure 2-3 shows a better way to represent the same information shown in Figure 2-2. In Figure 2-3, every position in the table contains a single value.
Database Design Fundamentals
ORDERS 28
ORDER_ ORDER_ NUM DATE
CUSTOMER_ PART_ NUM_ QUOTED_ NUM NUM ORDERED PRICE
21608
10/20/2010
148
AT94
11
$21.95
21610
10/20/2010
356
DR93
1
$495.00
21610
10/20/2010
356
DW11
1
$399.99
21613
10/21/2010
408
KL62
4
$329.95
21614
10/21/2010
282
KT03
2
$595.00
21617
10/23/2010
608
BV06
2
$12.95
21617
10/23/2010
608
CD52
4
$150.00
21619
10/23/2010
148
DR93
1
$495.00
21623
10/23/2010
608
KV29
2
$325.99
FIGURE 2-3
ORDERS data without repeating groups
When you remove the repeating groups from Figure 2-2, all of the rows in Figure 2-3 are single-valued. This structure is formally called a relation. A relation is a twodimensional table in which the entries in the table are single-valued (each location in the table contains a single entry), each column has a distinct name, all values in the column match this name, the order of the rows and columns is immaterial, and each row contains unique values. A relational database is a collection of relations.
NOTE Rows in a table (relation) are also called records or tuples. Columns in a table (relation) are also called fields or attributes. This text uses the terms tables, columns, and rows unless the more formal terms of relation, attributes, and tuples are necessary for clarity.
There is a commonly accepted shorthand representation to show the tables and columns in a relational database: for each table, you write the name of the table and then within parentheses list all of the columns in the table. In this representation, each table appears on its own line. Using this method, you represent the Premiere Products database as follows: REP (REP_NUM, LAST_NAME, FIRST_NAME, STREET, CITY, STATE, ZIP, COMMISSION, RATE) CUSTOMER (CUSTOMER_NUM, CUSTOMER_NAME, STREET, CITY, STATE, ZIP, BALANCE, CREDIT_LIMIT, REP_NUM) ORDERS (ORDER_NUM, ORDER_DATE, CUSTOMER_NUM) ORDER_LINE (ORDER_NUM, PART_NUM, NUM_ORDERED, QUOTED_PRICE) PART (PART_NUM, DESCRIPTION, ON_HAND, CLASS, WAREHOUSE, PRICE) Chapter 2
Notice that some tables contain columns with duplicate names. For example, the REP_NUM column appears in both the REP table and the CUSTOMER table. Suppose a situation existed wherein someone (or the DBMS) might confuse the two columns. For example, if you write REP_NUM, it is not clear which REP_NUM column you want to use. You need a mechanism for indicating the REP_NUM column to which you are referring. One common approach to solving this problem is to write both the table name and the column name, separated by a period. Thus, you would reference the REP_NUM column in the CUSTOMER table as CUSTOMER.REP_NUM, and the REP_NUM column in the REP table as REP.REP_NUM. Technically, when you reference columns in this format, you say that you qualify the names. It is always acceptable to qualify column names, even when there is no potential for confusion. If confusion might arise, however, it is essential to qualify column names.
29
FUNCTIONAL DEPENDENCE The concept of functional dependence is crucial to understanding the rest of the material in this chapter. Functional dependence is a formal name for what is basically a simple idea. To illustrate functional dependence, suppose the REP table for Premiere Products is structured as shown in Figure 2-4. The only difference between the REP table shown in Figure 2-4 and the one shown in Figure 2-1 is the addition of an extra column named PAY_CLASS. REP REP_ LAST_ FIRST_ NUM NAME NAME
STREET
CITY
20
Kaiser Valerie
624 Randall Grove
35 65
STATE ZIP
COMMISSION PAY_ RATE CLASS
FL
33321
$20,542.50 1
0.05
Hull
Richard 532 Jackson Sheldon FL
33553
$39,216.00 2
0.07
Perez
Juan
33336
$23,487.00 1
0.05
FIGURE 2-4
1626 Taylor Fillmore FL
REP table with a PAY_CLASS column
Suppose one of the policies at Premiere Products is that all sales reps in any given pay class earn their commissions at the same rate. To describe this situation, you could say that a sales rep’s pay class determines his or her commission rate. Alternatively, you could say that a sales rep’s commission rate depends on his or her pay class. This phrasing uses the words determines and depends on in the same way that you describe functional dependency. If you wanted to be formal, you would precede either expression with the word functionally. For example, you might say, “A sales rep’s pay class functionally determines his or her commission rate,” and “A sales rep’s commission rate functionally depends on his or her pay class.” You can also define functional dependency by saying that when you know a sales rep’s pay class, you can determine his or her commission rate. In a relational database, column B is functionally dependent on another column (or a collection of columns), A, if at any point in time a value for A determines a single value for B. You can think of this as follows: when you are given a value for A, do you know that
Database Design Fundamentals
30
you can find a single value for B? If so, B is functionally dependent on A (often written as A B). If B is functionally dependent on A, you also can say that A functionally determines B. At Premiere Products, is the LAST_NAME column in the REP table functionally dependent on the REP_NUM column? Yes, it is. If you are given a value for REP_NUM, such as 20, there is a single LAST_NAME, Kaiser, associated with it. This is represented as: REP_NUM
LAST_NAME
Q&A Question: In the CUSTOMER table, is CUSTOMER_NAME functionally dependent on REP_NUM? Answer: No. Given the REP_NUM 20, for example, you would not be able to find a single customer name, because 20 appears on more than one row in the table.
Q&A Question: In the ORDER_LINE table, is NUM_ORDERED functionally dependent on ORDER_NUM? Answer: No. An ORDER_NUM might be associated with several items in an order, so having just an ORDER_NUM does not provide enough information.
Q&A Question: Is NUM_ORDERED functionally dependent on PART_NUM? Answer: No. Again, just as with ORDER_NUM, a PART_NUM might be associated with several items in an order, so PART_NUM does not provide enough information.
Q&A Question: On which columns in the ORDER_LINE table is NUM_ORDERED functionally dependent? Answer: To determine a value for NUM_ORDERED, you need both an order number and a part number. In other words, NUM_ORDERED is functionally dependent on the combination (formally called the concatenation) of ORDER_NUM and PART_NUM. That is, given an order number and a part number, you can find a single value for NUM_ORDERED.
Chapter 2
At this point, a question naturally arises: how do you determine functional dependencies? Can you determine them by looking at sample data, for example? The answer is no. Consider the REP table in Figure 2-5, in which last names are unique. It is very tempting to say that LAST_NAME functionally determines STREET, CITY, STATE, and ZIP (or equivalently that STREET, CITY, STATE, and ZIP are all functionally dependent on LAST_NAME). After all, given the last name of a rep, you can find the single address.
31
REP REP_ LAST_ FIRST_ NUM NAME NAME
STREET
CITY
20
Kaiser Valerie
624 Randall Grove
35 65
STATE ZIP
COMMISSION RATE
FL
33321
$20,542.50
0.05
Hull
Richard 532 Jackson Sheldon FL
33553
$39,216.00
0.07
Perez
Juan
33336
$23,487.00
0.05
FIGURE 2-5
1626 Taylor Fillmore FL
REP table
What happens when rep 85, whose last name is also Kaiser, is added to the database? You then have the situation illustrated in Figure 2-6. Because there are now two reps with the last name of Kaiser, you can no longer find a single address using a rep’s last name—you were misled by the original data. The only way to determine functional dependencies is to examine the user’s policies. This process can involve discussions with users, an examination of user documentation, and so on. For example, if managers at Premiere Products have a policy never to hire two reps with the same last name, then LAST_NAME would indeed determine the other columns. Without such a policy, however, LAST_NAME would not determine the other columns. REP REP_ LAST_ FIRST_ NUM NAME NAME
STREET
CITY
20
Kaiser Valerie
624 Randall Grove
35
STATE ZIP
COMMISSION RATE
FL
33321
$20,542.50
0.05
Hull
Richard 532 Jackson Sheldon FL
33553
$39,216.00
0.07
65
Perez
Juan
33336
$23,487.00
0.05
85
Kaiser William 172 Bahia
39281
$0.00
0.05
FIGURE 2-6
1626 Taylor Fillmore FL Norton
FL
REP table with two reps named Kaiser
PRIMARY KEYS Another important database design concept is the primary key. In the simplest terms, the primary key is the unique identifier for a table. For example, the REP_NUM column is the unique identifier for the REP table. Given a rep number in the table, such as 20, there Database Design Fundamentals
32
will only be one row on which that rep number occurs. Thus, the rep number 20 uniquely identifies a row (in this case, the first row, and the rep named Valerie Kaiser). In this text, the definition of primary key needs to be more precise than a unique identifier for a table. Specifically, column A (or a collection of columns) is the primary key for a table if: Property 1. All columns in the table are functionally dependent on A. Property 2. No subcollection of the columns in A (assuming A is a collection of columns and not just a single column) also has property 1.
Q&A Question: Is the CLASS column the primary key for the PART table? Answer: No, because the other columns are not functionally dependent on CLASS. Given the class HW, for example, you cannot determine a part number, description, or anything else, because there are several rows on which the class is HW.
Q&A Question: Is the CUSTOMER_NUM column the primary key for the CUSTOMER table? Answer: Yes, because Premiere Products assigns unique customer numbers. A specific customer number cannot appear on more than one row. Thus, all columns in the CUSTOMER table are functionally dependent on CUSTOMER_NUM.
Q&A Question: Is the ORDER_NUM column the primary key for the ORDER_LINE table? Answer: No, because it does not functionally determine either NUM_ORDERED or QUOTED_PRICE.
Q&A Question: Is the combination of the ORDER_NUM and PART_NUM columns the primary key for the ORDER_LINE table? Answer: Yes, because you can determine all columns by this combination of columns, and, further, neither the ORDER_NUM nor the PART_NUM alone has this property.
Chapter 2
Q&A Question: Is the combination of the PART_NUM and DESCRIPTION columns the primary key for the PART table? Answer: No. Although it is true that you can determine all columns in the PART table by this combination, PART_NUM alone also has this property.
33
You can indicate a table’s primary key with a shorthand representation of a database by underlining the column or collection of columns that comprise the primary key. The complete shorthand representation for the Premiere Products database is: REP (REP_NUM, LAST_NAME, FIRST_NAME, STREET, CITY, STATE, ZIP, COMMISSION, RATE) CUSTOMER (CUSTOMER_NUM, CUSTOMER_NAME, STREET, CITY, STATE, ZIP, BALANCE, CREDIT_LIMIT, REP_NUM) ORDERS (ORDER_NUM, ORDER_DATE, CUSTOMER_NUM) ORDER_LINE (ORDER_NUM, PART_NUM, NUM_ORDERED, QUOTED_PRICE) PART (PART_NUM, DESCRIPTION, ON_HAND, CLASS, WAREHOUSE, PRICE)
NOTE Sometimes you might identify one or more columns that you can use as a table’s primary key. For example, if the Premiere Products database also included an EMPLOYEE table that contains employee numbers and Social Security numbers, either the employee number or the Social Security number could serve as the table’s primary key. In this case, both columns are referred to as candidate keys. Like a primary key, a candidate key is a column or collection of columns on which all columns in the table are functionally dependent—the definition for primary key really defines candidate key as well. From all the candidate keys, you would choose one to be the primary key.
NOTE According to the definition of a candidate key, a Social Security number is a legitimate primary key. Many databases, such as those that store data about students at a college or university or those that store data about employees at a company, store a person’s Social Security number as a primary key. However, many institutions and organizations are moving away from using Social Security numbers as primary keys because of privacy issues. Instead of using Social Security numbers, many institutions and organizations use unique student numbers or employee numbers as primary keys.
Database Design Fundamentals
NOTE 34
Some institutions prefer to assign values to use as primary keys for items such as customer numbers, part numbers, and student numbers. Others simply let the computer generate the values. In this case, the DBMS simply assigns the next available value. For example, if the database has already assigned customer numbers 1000 through 1436, it assigns the next new customer added to the database the customer number 1437.
DATA BA S E D E S I G N This section presents a specific method you can follow to design a database when given a set of requirements that the database must support. The determination of the requirements is part of the process known as systems analysis. A systems analyst interviews users, examines existing and proposed documents, and examines organizational policies to determine exactly the type of data needs the database must support. This text does not cover this analysis. Rather, it focuses on how to take the set of requirements that this process produces and determine the appropriate database design. After presenting the database design method, this section presents a sample set of requirements and illustrates the design method by designing a database to satisfy these requirements.
Design Method To design a database for a set of requirements, complete the following steps: 1.
2.
3.
4.
Chapter 2
Read the requirements, identify the entities (objects) involved, and name the entities. For example, when the design involves departments and employees, you might use the entity names DEPARTMENT and EMPLOYEE. When the design involves customers and sales reps, you might use the entity names CUSTOMER and REP. Identify the unique identifiers for the entities you identified in Step 1. For example, when one of the entities is PART, determine what information is required to uniquely identify each individual part. In other words, what information does the organization use to distinguish one part from another? For a PART entity, the unique identifier for each part might be a PART_NUM; for a CUSTOMER entity, the unique identifier might be a CUSTOMER_NUM. When no unique identifier is available from the data you know about the entity, you need to create one. For example, you might use a unique number to identify parts when no part numbers exist. Identify the attributes for all the entities. These attributes become the columns in the tables. It is possible for two or more entities to contain the same attributes. At Premiere Products, for example, reps and customers both have addresses, cities, states, and zip codes. To clarify this duplication of attributes, follow the name of the attribute with the corresponding entity in parentheses. Thus, ADDRESS (CUSTOMER) is a customer address and ADDRESS (REP) is a sales rep address. Identify the functional dependencies that exist among the attributes. Ask yourself the following question: if you know a unique value for an attribute, do you also
5.
6.
know the unique values for other attributes? For example, when you have the three attributes REP_NUM, LAST_NAME, and FIRST_NAME and you know a unique value for REP_NUM, do you also know a unique value for LAST_NAME and FIRST_NAME? If so, then LAST_NAME and FIRST_NAME are functionally dependent on REP_NUM (REP_NUM LAST_NAME, FIRST_NAME). Use the functional dependencies to identify the tables by placing each attribute with the attribute or minimum combination of attributes on which it is functionally dependent. The attribute or attributes for an entity on which all other attributes are dependent will be the primary key of the table. The remaining attributes will be the other columns in the table. Once you have determined all the columns in the table, you can give the table an appropriate name. Usually the name will be the same as the name you identified for the entity in Step 1. Identify any relationships between tables. In some cases, you might be able to determine the relationships directly from the requirements. It might be clear, for example, that one rep is related to many customers and that each customer is related to exactly one rep. When it is not, look for matching columns in the tables you created. For example, if both the REP table and the CUSTOMER table contain a REP_NUM column and the values in these columns must match, you know that reps and customers are related. The fact that the REP_NUM column is the primary key in the REP table tells you that the REP table is the “one” part of the relationship and the CUSTOMER table is the “many” part of the relationship.
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In the next section, you will apply this process to produce the design for the Premiere Products database using the collection of requirements that this database must support.
Database Design Requirements The analyst has interviewed users and examined documents at Premiere Products and has determined that the database must support the following requirements: 1. 2.
3.
4.
For a sales rep, store the sales rep’s number, last name, first name, street address, city, state, zip code, total commission, and commission rate. For a customer, store the customer’s number, name, street address, city, state, zip code, balance, and credit limit. In addition, store the number, last name, and first name of the sales rep who represents this customer. The analyst has also determined that a sales rep can represent many customers, but a customer must have exactly one sales rep (in other words, a sales rep must represent a customer; a customer cannot be represented by zero or more than one sales reps). For a part, store the part’s number, description, units on hand, item class, the number of the warehouse in which the part is located, and the price. All units of a particular part are stored in the same warehouse. For an order, store the order number, order date, the number and name of the customer that placed the order, and the number of the sales rep who represents that customer. Database Design Fundamentals
5. 36
For each line item within an order, store the part number and description, the number ordered, and the quoted price. The analyst also obtained the following information concerning orders: a. There is only one customer per order. b. On a given order, there is at most one line item for a given part. For example, part DR93 cannot appear on several lines within the same order. c. The quoted price might differ from the actual price when the sales rep discounts a certain part on a specific order.
Database Design Process Example The following steps apply the design process to the requirements for Premiere Products to produce the appropriate database design: Step 1: There appear to be four entities: reps, customers, parts, and orders. The names assigned to these entities are REP, CUSTOMER, PART, and ORDERS, respectively. Step 2: From the collection of entities, review the data and determine the unique identifier for each entity. For the REP, CUSTOMER, PART, and ORDERS entities, the unique identifiers are the rep number, customer number, part number, and order number, respectively. These unique identifiers are named REP_NUM, CUSTOMER_NUM, PART_NUM, and ORDER_NUM, respectively. Step 3: The attributes mentioned in the first requirement all refer to sales reps. The specific attributes mentioned in the requirement are the sales rep’s number, name, street address, city, state, zip code, total commission, and commission rate. Assigning appropriate names to these attributes produces the following list: REP_NUM LAST_NAME FIRST_NAME STREET CITY STATE ZIP COMMISSION RATE
The attributes mentioned in the second requirement refer to customers. The specific attributes are the customer’s number, name, street address, city, state, zip code, balance, and credit limit. The requirement also mentions the number, first name, and last name of the sales rep who represents this customer. Assigning appropriate names to these attributes produces the following list: CUSTOMER_NUM CUSTOMER_NAME STREET CITY STATE ZIP BALANCE CREDIT_LIMIT REP_NUM LAST_NAME FIRST_NAME
Chapter 2
There are attributes named STREET, CITY, STATE, and ZIP for sales reps as well as attributes named STREET, CITY, STATE, and ZIP for customers. To distinguish these attributes in the final collection, follow the name of the attribute by the name of the corresponding entity. For example, the street for a sales rep is STREET (REP) and the street for a customer is STREET (CUSTOMER). The attributes mentioned in the third requirement refer to parts. The specific attributes are the part’s number, description, units on hand, item class, the number of the warehouse in which the part is located, and the price. Assigning appropriate names to these attributes produces the following list:
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PART_NUM DESCRIPTION ON_HAND CLASS WAREHOUSE PRICE
The attributes mentioned in the fourth requirement refer to orders. The specific attributes include the order number, order date, number and name of the customer that placed the order, and number of the sales rep who represents the customer. Assigning appropriate names to these attributes produces the following list: ORDER_NUM ORDER_DATE CUSTOMER_NUM CUSTOMER_NAME REP_NUM
The specific attributes associated with the statement in the requirements concerning line items are the order number (to determine the order to which the line item corresponds), part number, description, number ordered, and quoted price. If the quoted price must be the same as the price, you could simply call it PRICE. According to requirement 5c, however, the quoted price might differ from the price, so you must add the quoted price to the list. Assigning appropriate names to these attributes produces the following list: ORDER_NUM PART_NUM DESCRIPTION NUM_ORDERED QUOTED_PRICE
The complete list grouped by entity is as follows: REP REP_NUM LAST_NAME FIRST_NAME STREET (REP) CITY (REP) STATE (REP) ZIP (REP) COMMISSION RATE
Database Design Fundamentals
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CUSTOMER CUSTOMER_NUM CUSTOMER_NAME STREET (CUSTOMER) CITY (CUSTOMER) STATE (CUSTOMER) ZIP (CUSTOMER) BALANCE CREDIT_LIMIT REP_NUM LAST_NAME FIRST_NAME PART PART_NUM DESCRIPTION ON_HAND CLASS WAREHOUSE PRICE ORDER ORDER_NUM ORDER_DATE CUSTOMER_NUM CUSTOMER_NAME REP_NUM For line items within an order ORDER_NUM PART_NUM DESCRIPTION NUM_ORDERED QUOTED_PRICE
Step 4: The fact that the unique identifier for sales reps is the rep number gives the following functional dependencies: REP_NUM LAST_NAME, FIRST_NAME, STREET (REP), CITY (REP), STATE (REP), ZIP (REP), COMMISSION, RATE
This notation indicates that the LAST_NAME, FIRST_NAME, STREET (REP), CITY (REP), STATE (REP), ZIP (REP), COMMISSION, and RATE are all functionally dependent on REP_NUM. The fact that the unique identifier for customers is the customer number gives the following functional dependencies: CUSTOMER_NUM CUSTOMER_NAME, STREET (CUSTOMER), CITY (CUSTOMER), STATE (CUSTOMER), ZIP (CUSTOMER), BALANCE, CREDIT_LIMIT, REP_NUM, LAST_NAME, FIRST_NAME
Chapter 2
Q&A Question: Do you really need to include the last name and first name of a sales rep in the list of attributes determined by the customer number? Answer: There is no need to include them in this list, because they both can be determined from the sales rep number and are already included in the list of attributes determined by REP_NUM.
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Thus, the functional dependencies for the CUSTOMER entity are as follows: CUSTOMER_NUM CUSTOMER_NAME, STREET (CUSTOMER), CITY (CUSTOMER), STATE (CUSTOMER), ZIP (CUSTOMER), BALANCE, CREDIT_LIMIT, REP_NUM
The fact that the unique identifier for parts is the part number gives the following functional dependencies: PART_NUM
DESCRIPTION, ON_HAND, CLASS, WAREHOUSE, PRICE
The fact that the unique identifier for orders is the order number gives the following functional dependencies: ORDER_DATE, CUSTOMER_NUM, CUSTOMER_NAME, ORDER_NUM REP_NUM
Q&A Question: Do you really need to include the name of a customer and the number of the customer’s rep in the list of attributes determined by the order number? Answer: There is no need to include the customer name and the rep number in this list, because you can determine them from the customer number and they are already included in the list of attributes determined by CUSTOMER_NUM.
The functional dependencies for the ORDERS entity are as follows: ORDER_NUM
ORDER_DATE, CUSTOMER_NUM
The final attributes to be examined are those associated with the line items within the order: PART_NUM, DESCRIPTION, NUM_ORDERED, and QUOTED_PRICE.
Database Design Fundamentals
Q&A 40
Question: Why aren’t NUM_ORDERED and QUOTED_PRICE included in the list of attributes determined by the order number? Answer: To uniquely identify a particular value for NUM_ORDERED or QUOTED_PRICE, ORDER_NUM alone is not sufficient. It requires the combination of ORDER_NUM and PART_NUM.
The following shorthand representation indicates that the combination of ORDER_NUM and PART_NUM functionally determines NUM_ORDERED and QUOTED_PRICE: ORDER_NUM, PART_NUM
NUM_ORDERED, QUOTED_PRICE
Q&A Question: Does DESCRIPTION need to be included in this list? Answer: No, because DESCRIPTION can be determined by the PART_NUMBER alone, and it already appears in the list of attributes dependent on the PART_NUM.
The complete list of functional dependencies is as follows: REP_NUM LAST_NAME, FIRST_NAME, STREET (REP), CITY (REP), STATE (REP), ZIP(REP), COMMISSION, RATE CUSTOMER_NAME, STREET (CUSTOMER), CUSTOMER_NUM CITY (CUSTOMER), STATE (CUSTOMER), ZIP (CUSTOMER), BALANCE, CREDIT_LIMIT, REP_NUM DESCRIPTION, ON_HAND, CLASS, WAREHOUSE, PRICE PART_NUM ORDER_DATE, CUSTOMER_NUM ORDER_NUM NUM_ORDERED, QUOTED_PRICE ORDER_NUM, PART_NUM
Step 5: Using the functional dependencies, you can create tables with the attribute(s) to the left of the arrow being the primary key and the items to the right of the arrow being the other columns. For relations corresponding to those entities identified in Step 1, you can use the name you already determined. Because you did not identify any entity that had a unique identifier that was the combination of ORDER_NUM and PART_NUM, you need to assign a name to the table whose primary key consists of these two columns. Because this table represents the individual lines within an order, the name ORDER_LINE is a good choice. The final collection of tables is as follows: REP (REP_NUM, LAST_NAME, FIRST_NAME, STREET, CITY, STATE, ZIP, COMMISSION, RATE) CUSTOMER (CUSTOMER_NUM, CUSTOMER_NAME, STREET, CITY, STATE, ZIP, BALANCE, CREDIT_LIMIT, REP_NUM) PART (PART_NUM, DESCRIPTION, ON_HAND, CLASS, WAREHOUSE, PRICE) ORDERS (ORDER_NUM, ORDER_DATE, CUSTOMER_NUM) ORDER_LINE (ORDER_NUM, PART_NUM, NUM_ORDERED, QUOTED_PRICE)
Chapter 2
Step 6: Examining the tables and identifying common columns gives the following list of relationships between the tables: ●
●
●
●
The CUSTOMER and REP tables are related using the REP_NUM columns. Because the REP_NUM column is the primary key for the REP table, this indicates a one-to-many relationship between REP and CUSTOMER (one rep to many customers). The ORDERS and CUSTOMER tables are related using the CUSTOMER_NUM columns. Because the CUSTOMER_NUM column is the primary key for the CUSTOMER table, this indicates a one-to-many relationship between CUSTOMER and ORDERS (one customer to many orders). The ORDER_LINE and ORDERS tables are related using the ORDER_NUM columns. Because the ORDER_NUM column is the primary key for the ORDERS table, this indicates a one-to-many relationship between ORDERS and ORDER_LINE (one order to many order lines). The ORDER_LINE and PART tables are related using the PART_NUM columns. Because the PART_NUM column is the primary key for the PART table, this indicates a one-to-many relationship between PART and ORDER_LINE (one part to many order lines).
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N O R M A L I Z AT I O N After creating the database design, you must analyze it to make sure it is free of potential problems. To do so, you follow a process called normalization, in which you identify the existence of potential problems, such as data duplication and redundancy, and implement ways to correct these problems. The goal of normalization is to convert unnormalized relations (tables that satisfy the definition of a relation except that they might contain repeating groups) into various types of normal forms. A table in a particular normal form possesses a certain desirable collection of properties. Although there are several normal forms, the most common are first normal form, second normal form, and third normal form. Normalization is a process in which a table that is in first normal form is better than a table that is not in first normal form, a table that is in second normal form is better than one that is in first normal form, and so on. The goal of this process is to allow you to take a table or collection of tables and produce a new collection of tables that represents the same information but is free of problems.
First Normal Form According to the definition of a relation, a relation (table) cannot contain a repeating group in which multiple entries exist on a single row. However, in the database design process, you might create a table that has all the other properties of a relation, but contains a repeating group. Removing repeating groups is the starting point when converting an unnormalized collection of data into a table that is in first normal form. A table (relation) is in first normal form (1NF) when it does not contain a repeating group.
Database Design Fundamentals
42
For example, in the design process you might create the following ORDERS table, in which there is a repeating group consisting of PART_NUM and NUM_ORDERED. The notation for this table is as follows: ORDERS (ORDER_NUM, ORDER_DATE, (PART_NUM, NUM_ORDERED) )
This notation describes a table named ORDERS that consists of a primary key, ORDER_NUM, and a column named ORDER_DATE. The inner parentheses indicate a repeating group that contains two columns, PART_NUM and NUM_ORDERED. This table contains one row per order with values in the PART_NUM and NUM_ORDERED columns for each order with the number ORDER_NUM and placed on ORDER_DATE. Figure 2-7 shows a single order with multiple combinations of a part number and a corresponding number of units ordered. ORDERS ORDER_ ORDER_ NUM DATE
PART_ NUM
NUM_ ORDERED
21608
10/20/2010 AT94
11
21610
10/20/2010 DR93 DW11
1 1
21613
10/21/2010 KL62
4
21614
10/21/2010 KT03
2
21617
10/23/2010 BV06 CD52
2 4
21619
10/23/2010 DR93
1
21623
10/23/2010 KV29
2
FIGURE 2-7
Unnormalized order data
To convert the table to first normal form, you remove the repeating group as follows: ORDERS (ORDER_NUM, ORDER_DATE, PART_NUM, NUM_ORDERED)
Chapter 2
Figure 2-8 shows the table in first normal form. 43
ORDERS ORDER_ ORDER_ DATE NUM
PART_ NUM_ NUM ORDERED
21608
10/20/2010 AT94
11
21610
10/20/2010 DR93
1
21610
10/20/2010 DW11
1
21613
10/21/2010 KL62
4
21614
10/21/2010 KT03
2
21617
10/23/2010 BV06
2
21617
10/23/2010 CD52
4
21619
10/23/2010 DR93
1
21623
10/23/2010 KV29
2
FIGURE 2-8
Order data converted to first normal form
In Figure 2-7, the second row indicates that part DR93 and part DW11 are both included in order 21610. In Figure 2-8, this information is represented by two rows, the second and third. The primary key for the unnormalized ORDERS table was the ORDER_NUM column alone. The primary key for the normalized table is now the combination of the ORDER_NUM and PART_NUM columns. When you convert an unnormalized table to a table in first normal form, the primary key of the table in first normal form is usually the primary key of the unnormalized table concatenated with the key for the repeating group, which is the column in the repeating group that distinguishes one occurrence of the repeating group from another within a given row in the table. In the ORDERS table, PART_NUM was the key to the repeating group and ORDER_NUM was the primary key for the table. When converting the unnormalized data to first normal form, the primary key becomes the concatenation of the ORDER_NUM and PART_NUM columns.
Second Normal Form The following ORDERS table is in first normal form, because it does not contain a repeating group: ORDERS (ORDER_NUM, ORDER_DATE, PART_NUM, DESCRIPTION, NUM_ORDERED, QUOTED_PRICE)
The table contains the following functional dependencies: ORDER_NUM ORDER_DATE DESCRIPTION PART_NUM NUM_ORDERED, QUOTED_PRICE ORDER_NUM, PART_NUM
Database Design Fundamentals
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This notation indicates that ORDER_NUM alone determines ORDER_DATE, and PART_NUM alone determines DESCRIPTION, but it requires both an ORDER_NUM and a PART_NUM to determine either NUM_ORDERED or QUOTED_PRICE. Consider the sample of this table shown in Figure 2-9. ORDERS ORDER_ ORDER_ DATE NUM
PART_ DESCRIPTION NUM
NUM_ ORDERED
QUOTED_ PRICE
11
$21.95
21608
10/20/2010 AT94
Iron
21610
10/20/2010 DR93
Gas Range
1
$495.00
21610
10/20/2010 DW11
Washer
1
$399.99
21613
10/21/2010 KL62
Dryer
4
$329.95
21614
10/21/2010 KT03
Dishwasher
2
$595.00
21617
10/23/2010 BV06
Home Gym
2
$12.95
21617
10/23/2010 CD52
Microwave Oven
4
$150.00
21619
10/23/2010 DR93
Gas Range
1
$495.00
21623
10/23/2010 KV29
Treadmill
2
$325.99
FIGURE 2-9
Sample ORDERS table
Although the ORDERS table is in first normal form (because it contains no repeating groups), problems exist within the table that require you to restructure it. The description of a specific part, DR93 for example, occurs twice in the table. This duplication (formally called redundancy) causes several problems. It is certainly wasteful of space, but that is not nearly as serious as some of the other problems. These other problems are called update anomalies and they fall into four categories: 1.
2.
3.
Chapter 2
Updates: If you need to change to the description of part DR93, you must change it twice—once in each row on which part DR93 appears. Updating the part description more than once makes the update process much more cumbersome and time consuming. Inconsistent data: There is nothing about the design that prohibits part DR93 from having two different descriptions in the database. In fact, if part DR93 occurs on 20 rows in the table, it is possible for this part to have 20 different descriptions in the database. Additions: When you try to add a new part and its description to the database, you will face a real problem. Because the primary key for the ORDERS table consists of both an ORDER_NUM and a PART_NUM, you need values for both of these columns to add a new row to the table. If you add a part to the table that does not yet have any orders, what do you use for an ORDER_NUM? The only solution is to create a dummy ORDER_NUM and then replace
4.
it with a real ORDER_NUM once an order for this part is actually received. Certainly this is not an acceptable solution. Deletions: If you delete order 21608 from the database and it is the only order that contains part AT94, deleting the order also deletes all information about part AT94. For example, you would no longer know that part AT94 is an iron.
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These problems occur because you have a column, DESCRIPTION, that is dependent on only a portion of the primary key, PART_NUM, and not on the complete primary key. This situation leads to the definition of second normal form. Second normal form represents an improvement over first normal form because it eliminates update anomalies in these situations. A table (relation) is in second normal form (2NF) when it is in first normal form and no nonkey column (that is, a column that is not part of the primary key) is dependent on only a portion of the primary key.
NOTE When the primary key of a table contains only a single column, the table is automatically in second normal form.
You can identify the fundamental problem with the ORDERS table: it is not in second normal form. Although it is important to identify the problem, what you really need is a method to correct it; you want to be able to convert tables to second normal form. First, take each subset of the set of columns that make up the primary key, and begin a new table with this subset as its primary key. For the ORDERS table, the new design is: (ORDER_NUM, (PART_NUM, (ORDER_NUM, PART_NUM,
Next, place each of the other columns with the appropriate primary key; that is, place each one with the minimal collection of columns on which it depends. For the ORDERS table, add the new columns as follows: (ORDER_NUM, ORDER_DATE) (PART_NUM, DESCRIPTION) (ORDER_NUM, PART_NUM, NUM_ORDERED, QUOTED_PRICE)
Each of these new tables is given a descriptive name based on the meaning and contents of the table, such as ORDERS, PART, and ORDER_LINE. Figure 2-10 shows samples of these tables.
Database Design Fundamentals
ORDERS 46
ORDER_ ORDER_ NUM DATE 21608 21610 21610 21613 21614 21617 21617 21619 21623
10/20/2010 10/20/2010 10/20/2010 10/21/2010 10/21/2010 10/23/2010 10/23/2010 10/23/2010 10/23/2010
ORDERS
FIGURE 2-10
Chapter 2
AT94 DR93 DW11 KL62 KT03 BV06 CD52 DR93 KV29
NUM_ QUOTED_ ORDERED PRICE
Iron Gas Range Washer Dryer Dishwasher Home Gym Microwave Oven Gas Range Treadmill
PART
ORDER_ ORDER_ NUM DATE 21608 21610 21613 21614 21617 21619 21623
PART_ DESCRIPTION NUM
10/20/2010 10/20/2010 10/21/2010 10/21/2010 10/23/2010 10/23/2010 10/23/2010
11 1 1 4 2 2 4 1 2
$21.95 $495.00 $399.99 $329.95 $595.00 $12.95 $150.00 $495.00 $325.99
ORDER_LINE
PART_ DESCRIPTION NUM AT94
Iron
BV06
Home Gym
CD52
Microwave Oven
DL71
Cordless Drill
DR93
Gas Range
DW11
Washer
FD21
Stand Mixer
KL62
Dryer
KT03 KV29
Dishwasher Treadmill
ORDER_ NUM
NUM_ QUOTED_ PART_ ORDERED PRICE NUM
21608 21610 21610 21613 21614 21617 21617 21619 21623
AT94 DR93 DW11 KL62 KT03 BV06 CD52 DR93 KV29
ORDERS table converted to second normal form
11 1 1 4 2 2 4 1 2
$21.95 $495.00 $399.99 $329.95 $595.00 $12.95 $150.00 $495.00 $325.99
In Figure 2-10, converting the original ORDERS table to a new ORDERS table, a PART table, and an ORDER_LINE table eliminates the update anomalies. A description appears only once for each part, so you do not have the redundancy that existed in the original table design. Changing the description of part DR93 from Gas Range to Deluxe Range, for example, is now a simple process involving a single change. Because the description for a part occurs in a single place, it is not possible to have multiple descriptions for a single part in the database at the same time. To add a new part and its description, you create a new row in the PART table, regardless of whether that part has pending or actual orders. Also, deleting order 21608 does not delete part number AT94 from the database because it still exists in the PART table. Finally, you have not lost any information by converting the ORDERS table to second normal form. You can reconstruct the data in the original table from the data in the new tables.
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Third Normal Form Problems can still exist with tables that are in second normal form. For example, suppose that you create the following CUSTOMER table: CUSTOMER (CUSTOMER_NUM, CUSTOMER_NAME, BALANCE, CREDIT_LIMIT, REP_NUM, LAST_NAME, FIRST_NAME)
This table has the following functional dependencies: CUSTOMER_NUM CUSTOMER_NAME, BALANCE, CREDIT_LIMIT, REP_NUM, LAST_NAME, FIRST_NAME LAST_NAME, FIRST_NAME REP_NUM
CUSTOMER_NUM determines all the other columns. In addition, REP_NUM determines LAST_NAME and FIRST_NAME. When a table’s primary key is a single column, the table is automatically in second normal form. (If the table were not in second normal form, some column would be dependent on only a portion of the primary key, which is impossible when the primary key is just one column.) Thus, the CUSTOMER table is in second normal form. Although this table is in second normal form, Figure 2-11 shows that it still possesses update problems similar to those identified for the ORDERS table shown in Figure 2-9. In Figure 2-11, the sales rep name occurs many times in the table.
Database Design Fundamentals
CUSTOMER 48
CUSTOMER_ CUSTOMER_NAME NUM
BALANCE
148
Al’s Appliance and Sport
$6,550.00
282
Brookings Direct
356
Ferguson’s
$5,785.00
408
The Everything Shop
$5,285.25
462
Bargains Galore
$3,412.00 $10,000.00
524
Kline’s
608
Johnson’s Department Store
$2,106.00 $10,000.00
65 Perez
Juan
687
Lee’s Sport and Appliance
$2,851.00
$5,000.00
35 Hull
Richard
725
Deerfield’s Four Seasons
$248.00
$7,500.00
35 Hull
Richard
842
All Season
$8,221.00
$7,500.00
20 Kaiser Valerie
FIGURE 2-11
CREDIT_ REP_ LAST_ FIRST_ LIMIT NUM NAME NAME
$7,500.00
20 Kaiser Valerie 35 Hull
Richard
$7,500.00
65 Perez
Juan
$5,000.00
35 Hull
Richard
65 Perez
Juan
$431.50 $10,000.00
$12,762.00 $15,000.00
20 Kaiser Valerie
Sample CUSTOMER table
The redundancy of including a sales rep number and name in the CUSTOMER table results in the same set of problems that existed for the ORDERS table. In addition to the problem of wasted space, you have the following update anomalies: 1. 2.
3.
4.
Updates: Changing the sales rep name requires changes to multiple rows in the table. Inconsistent data: The design does not prohibit multiple iterations of sales rep names in the database. For example, a sales rep might represent 20 customers and his name might be entered 20 different ways in the table. Additions: To add sales rep 87 (Emily Daniels) to the database, she must represent at least one customer. If Emily does not yet represent any customers, you either cannot record the fact that her name is Emily Daniels or you must create a fictitious customer for her to represent until she represents an actual customer. Neither of these solutions is desirable. Deletions: If you delete all the customers of sales rep 35 from the database, you will also lose all information about sales rep 35.
These update anomalies are due to the fact that REP_NUM determines LAST_NAME and FIRST_NAME, but REP_NUM is not the primary key. As a result, the same REP_NUM and consequently the same LAST_NAME and FIRST_NAME can appear on many different rows. You have seen that tables in second normal form represent an improvement over tables in first normal form, but to eliminate problems with tables in second normal form, you need an even better strategy for creating tables. Third normal form provides that strategy.
Chapter 2
Before looking at third normal form, however, you need to become familiar with the special name that is given to any column that determines another column (like REP_NUM in the CUSTOMER table). Any column (or collection of columns) that determines another column is called a determinant. A table’s primary key is a determinant. In fact, by definition, any candidate key is a determinant. (Remember that a candidate key is a column or collection of columns that could function as the primary key.) In Figure 2-11, REP_NUM is a determinant, but it is not a candidate key, and that is the problem. A table is in third normal form (3NF) when it is in second normal form and the only determinants it contains are candidate keys.
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NOTE This text’s definition of third normal form is not the original definition. This more recent definition, which is preferable to the original, is often referred to as Boyce-Codd normal form (BCNF) when it is important to make a distinction between this definition and the original definition. This text does not make such a distinction but will take this to be the definition of third normal form.
Now you have identified the problem with the CUSTOMER table: it is not in third normal form. There are several steps for converting tables to third normal form. First, for each determinant that is not a candidate key, remove from the table the columns that depend on this determinant (but do not remove the determinant). Next, create a new table containing all the columns from the original table that depend on this determinant. Finally, make the determinant the primary key of this new table. In the CUSTOMER table, for example, remove LAST_NAME and FIRST_NAME because they depend on the determinant REP_NUM, which is not a candidate key. A new table is formed, consisting of REP_NUM as the primary key, and the columns LAST_NAME and FIRST_NAME, as follows: CUSTOMER (CUSTOMER_NUM, CUSTOMER_NAME, BALANCE, CREDIT_LIMIT, REP_NUM)
and REP (REP_NUM, LAST_NAME, FIRST_NAME)
Figure 2-12 shows the original CUSTOMER table and the tables created when converting the original table to third normal form.
Database Design Fundamentals
CUSTOMER 50
CUSTOMER_ NUM
CUSTOMER_NAME
BALANCE
148 282 356 408 462 524 608 687 725 842
Al’s Appliance and Sport Brookings Direct Ferguson’s The Everything Shop Bargains Galore Kline’s Johnson’s Department Store Lee’s Sport and Appliance Deerfield’s Four Seasons All Season
$6,550.00 $431.50 $5,785.00 $5,285.25 $3,412.00 $12,762.00 $2,106.00 $2,851.00 $248.00 $8,221.00
CREDIT_ REP_ LIMIT NUM $7,500.00 $10,000.00 $7,500.00 $5,000.00 $10,000.00 $15,000.00 $10,000.00 $5,000.00 $7,500.00 $7,500.00
20 35 65 35 65 20 65 35 35 20
LAST_ FIRST_ NAME NAME Kaiser Hull Perez Hull Perez Kaiser Perez Hull Hull Kaiser
Valerie Richard Juan Richard Juan Valerie Juan Richard Richard Valerie
CUSTOMER CUSTOMER_ CUSTOMER_NAME NUM 148 282 356 408 462 524 608 687 725 842
BALANCE CREDIT_LIMIT REP_NUM
Al’s Appliance and Sport $6,550.00 Brookings Direct $431.50 Ferguson’s $5,785.00 The Everything Shop $5,285.25 Bargains Galore $3,412.00 Kline’s $12,762.00 Johnson’s Department Store $2,106.00 Lee’s Sport and Appliance $2,851.00 Deerfield’s Four Seasons $248.00 All Season $8,221.00
$7,500.00 $10,000.00 $7,500.00 $5,000.00 $10,000.00 $15,000.00 $10,000.00 $5,000.00 $7,500.00 $7,500.00
20 35 65 35 65 20 65 35 35 20
REP REP_NUM 20 35 65
FIGURE 2-12
Chapter 2
CUSTOMER table converted to third normal form
LAST_NAME Kaiser Hull Perez
FIRST_NAME Valerie Richard Juan
Has this new design for the CUSTOMER table corrected all of the previously identified problems? A sales rep’s name appears only once, thus avoiding redundancy and simplifying the process of changing a sales rep’s name. This design prohibits a sales rep from having different names in the database. To add a new sales rep to the database, you add a row to the REP table; it is not necessary for a new rep to represent a customer. Finally, deleting all customers of a given sales rep will not remove the sales rep’s record from the REP table, retaining the sales rep’s name in the database. You can reconstruct all the data in the original table from the data in the new collection of tables. All previously mentioned problems have indeed been solved.
51
Q&A Question: Convert the following table to third normal form. In this table, STUDENT_NUM determines STUDENT_NAME, NUM_CREDITS, ADVISOR_NUM, and ADVISOR_NAME. ADVISOR_NUM determines ADVISOR_NAME. COURSE_NUM determines DESCRIPTION. The combination of a STUDENT_NUM and a COURSE_NUM determines GRADE. STUDENT (STUDENT_NUM, STUDENT_NAME, NUM_CREDITS, ADVISOR_NUM, ADVISOR_NAME, (COURSE_NUM, DESCRIPTION, GRADE) )
Answer: Complete the following steps: Step 1. Remove the repeating group to convert the table to first normal form, as follows: STUDENT (STUDENT_NUM, STUDENT_NAME, NUM_CREDITS, ADVISOR_NUM, ADVISOR_NAME, COURSE_NUM, DESCRIPTION, GRADE)
The STUDENT table is now in first normal form because it has no repeating groups. It is not, however, in second normal form because STUDENT_NAME is dependent only on STUDENT_NUM, which is only a portion of the primary key. Step 2. Convert the STUDENT table to second normal form. First, for each subset of the primary key, start a table with that subset as its key yielding the following: (STUDENT_NUM, (COURSE_NUM, (STUDENT_NUM, COURSE_NUM,
Next, place the rest of the columns with the smallest collection of columns on which they depend, as follows: (STUDENT_NUM, STUDENT_NAME, NUM_CREDITS, ADVISOR_NUM, ADVISOR_NAME) (COURSE_NUM, DESCRIPTION) (STUDENT_NUM, COURSE_NUM, GRADE)
Finally, assign names to each of the new tables: STUDENT (STUDENT_NUM, STUDENT_NAME, NUM_CREDITS, ADVISOR_NUM, ADVISOR_NAME) COURSE (COURSE_NUM, DESCRIPTION) STUDENT_COURSE (STUDENT_NUM, COURSE_NUM, GRADE)
These tables are all now in second normal form, and the COURSE and STUDENT_COURSE tables are also in third normal form. The STUDENT table is not in third normal form, however, because it contains a determinant (ADVISOR_NUM) that is not a candidate key. continued
Database Design Fundamentals
52
Step 3: Convert the STUDENT table to third normal form by removing the column that depends on the determinant ADVISOR_NUM and placing it in a separate table, as follows: (STUDENT_NUM, STUDENT_NAME, NUM_CREDITS, ADVISOR_NUM) (ADVISOR_NUM, ADVISOR_NAME)
Step 4: Name the tables and put the entire collection together, as follows: STUDENT (STUDENT_NUM, STUDENT_NAME, NUM_CREDITS, ADVISOR_NUM) ADVISOR (ADVISOR_NUM, ADVISOR_NAME) COURSE (COURSE_NUM, DESCRIPTION) STUDENT_COURSE (STUDENT_NUM, COURSE_NUM, GRADE)
D I AG R A M S F O R DATA BA S E D E S I G N For many people, an illustration of a database’s structure is quite useful. A popular type of illustration used to represent the structure of a database is the entity-relationship (E-R) diagram. In an E-R diagram, a rectangle represents an entity (table). One-to-many relationships between entities are drawn as lines between the corresponding rectangles. Several different styles of E-R diagrams are used to diagram a database design. In the version shown in Figure 2-13, an arrowhead indicates the “many” side of the relationship between tables. In the relationship between the REP and CUSTOMER tables, for example, the arrow points from the REP table to the CUSTOMER table, indicating that one sales rep is related to many customers. The ORDER_LINE table has two one-to-many relationships, as indicated by the line from the ORDERS table to the ORDER_LINE table and the line from the PART table to the ORDER_LINE table.
REP
Rectangle represents an entity
CUSTOMER
ORDERS
FIGURE 2-13
Chapter 2
Arrow represents a one-to-many relationship Arrowhead points to the “many” part of the relationship
ORDER_LINE
PART
E-R diagram for the Premiere Products database with rectangles and arrows
NOTE In this style of E-R diagram, you can put the rectangles in any position to represent the entities and relationships. The important thing is that the arrows connect the appropriate rectangles.
53
Another style of E-R diagram is to represent the “many” side of a relationship between tables with a crow’s foot, as shown in Figure 2-14.
REP
CUSTOMER
ORDERS
FIGURE 2-14
Crow’s foot represents the “many” part of the relationship
ORDER_LINE
PART
E-R diagram for the Premiere Products database with a crow’s foot
The E-R diagram shown in Figure 2-15 represents the original style of E-R diagrams. In this style, relationships are indicated in diamonds that describe the relationship. The relationship between the REP and CUSTOMER tables, for example, is named REPRESENTS, reflecting the fact that a sales rep represents a customer. The relationship between the CUSTOMER and ORDERS table is named PLACED, reflecting the fact that customers place orders. The relationship between the ORDERS and ORDER_LINE tables is named CONTAINS, reflecting the fact that an order contains order lines. The relationship between the PART and ORDER_LINE tables is named IS_ON, reflecting the fact that a given part is on many orders. In this style of E-R diagram, the number 1 indicates the “one” side of the relationship and the letter “n” represents the “many” side of the relationship.
Database Design Fundamentals
REP 54
“One” part of a relationship
1
Diamond represents and describes a relationship
REPRESENTS
“Many” part of a relationship
n
CUSTOMER 1
PLACED
n
ORDERS
FIGURE 2-15
Chapter 2
1
CONTAINS
n
ORDER_LINE
n
IS_ON
1
PART
E-R diagram for the Premiere Products database with named relationships
Chapter Summary ●
An entity is a person, place, thing, or event. An attribute is a property of an entity. A relationship is an association between entities.
●
A relation is a two-dimensional table in which the entries in the table contain only single values, each column has a distinct name, all values in a column match this name, the order of the rows and columns is immaterial, and each row contains unique values. A relational database is a collection of relations.
●
Column B is functionally dependent on another column, A (or possibly a collection of columns), when a value for A determines a single value for B at any one time.
●
Column A (or a collection of columns) is the primary key for a relation (table), R, if all columns in R are functionally dependent on A and no subcollection of the columns in A (assuming A is a collection of columns and not just a single column) also has property 1.
●
To design a database to satisfy a particular set of requirements, first read through the requirements and identify the entities (objects) involved. Give names to the entities and identify the unique identifiers for these entities. Next, identify the attributes for all the entities and the functional dependencies that exist among the attributes, and then use the functional dependencies to identify the tables and columns. Finally, identify any relationships between tables by looking at matching columns.
●
A table (relation) is in first normal form (1NF) when it does not contain a repeating group. To convert an unnormalized table to first normal form, remove the repeating group and expand the primary key to include the original primary key along with the key to the repeating group.
●
A table (relation) is in second normal form (2NF) when it is in first normal form and no nonkey column (that is, a column that is not part of the primary key) is dependent on only a portion of the primary key. To convert a table in first normal form to a collection of tables in second normal form, take each subset of the set of columns that make up the primary key, and begin a new table with this subset as its primary key. Next, place each of the other columns with the appropriate primary key; that is, place each one with the minimal collection of columns on which it depends. Finally, give each of these new tables a name that is descriptive of the meaning and contents of the table.
●
A table is in third normal form (3NF) when it is in second normal form and the only determinants (columns on which at least one other column depends) it contains are candidate keys (columns that could function as the primary key). To convert a table in second normal form to a collection of tables in third normal form, first, for each determinant that is not a candidate key, remove from the table the columns that depend on this determinant (but don’t remove the determinant). Next, create a new table containing all the columns from the original table that depend on this determinant. Finally, make the determinant the primary key of this new table.
●
An entity-relationship (E-R) diagram is an illustration that represents the design of a database. There are several common styles of illustrating database design that use shapes to represent entities and connectors to illustrate the relationships between those entities.
55
Database Design Fundamentals
Key Terms 56
attribute
one-to-many relationship
Boyce-Codd normal form (BCNF)
primary key
candidate key
qualify
concatenation
record
database design
redundancy
determinant
relation
entity
relational database
entity-relationship (E-R) diagram
relationship
field
repeating group
first normal form (1NF)
second normal form (2NF)
functionally dependent
third normal form (3NF)
functionally determine
tuple
nonkey column
unnormalized relation
normal form
update anomaly
normalization
Review Questions
Chapter 2
1.
What is an entity?
2.
What is an attribute?
3.
What is a relationship? What is a one-to-many relationship?
4.
What is a repeating group?
5.
What is a relation?
6.
What is a relational database?
7.
Describe the shorthand representation of the structure of a relational database. Illustrate this technique by representing the database for Henry Books as shown in Figures 1-4 through 1-7 in Chapter 1.
8.
How do you qualify the name of a field, and when do you need to do this?
9.
What does it mean for a column to be functionally dependent on another column?
10.
What is a primary key? What is the primary key for each of the tables in the Henry Books database shown in Chapter 1?
11.
A database at a college must support the following requirements: a.
For a department, store its number and name.
b.
For an advisor, store his or her number, last name, first name, and the department number to which the advisor is assigned.
c.
For a course, store its code and description (for example, MTH110, Algebra).
d.
For a student, store his or her number, first name, and last name. For each course the student takes, store the course code, the course description, and the grade earned.
Also, store the number and name of the student’s advisor. Assume that an advisor might advise any number of students but that each student has just one advisor. Design the database for the preceding set of requirements. Use your own experience as a student to determine any functional dependencies. List the tables, columns, and relationships. In addition, represent your design with an E-R diagram. 12.
Define first normal form.
13.
Define second normal form. What types of problems might you encounter using tables that are not in second normal form?
14.
Define third normal form. What types of problems might you encounter using tables that are not in third normal form?
15.
Using the functional dependencies you determined in Question 11, convert the following table to an equivalent collection of tables that are in third normal form.
57
STUDENT (STUDENT_NUM, STUDENT_LAST_NAME, STUDENT_FIRST_NAME, ADVISOR_NUM, ADVISOR_LAST_NAME, ADVISOR_FIRST_NAME, (COURSE_CODE, DESCRIPTION, GRADE) )
Exercises Premiere Products Answer each of the following questions using the Premiere Products data shown in Figure 2-1. No computer work is required. 1.
Indicate the changes (using the shorthand representation) that you would need to make to the original Premiere Products database design (see Figure 2-1) to support the following requirements. A customer is not necessarily represented by a single sales rep, but can be represented by several sales reps. When a customer places an order, the sales rep who gets the commission on the order must be in the collection of sales reps who represent the customer.
2.
Indicate the changes (using the shorthand representation) that you would need to make to the original Premiere Products database design to support the following requirements. There is no relationship between customers and sales reps. When a customer places an order, any sales rep can process the order. On the order, you need to identify both the customer placing the order and the sales rep responsible for the order. Draw an E-R diagram for the new design.
3.
Indicate the changes (using the shorthand representation) that you would need to make to the original Premiere Products database design in the event that the original Requirement 3 is changed as follows. For a part, store the part’s number, description, item class, and price. In addition, for each warehouse in which the part is located, store the number of the warehouse, the description of the warehouse, and the number of units of the part stored in the warehouse. Draw an E-R diagram for the new design.
Database Design Fundamentals
4. 58
Using your knowledge of Premiere Products, determine the functional dependencies that exist in the following table. After determining the functional dependencies, convert this table to an equivalent collection of tables that are in third normal form. PART (PART_NUM, DESCRIPTION, ON_HAND, CLASS, WAREHOUSE, PRICE, (ORDER_NUM, ORDER_DATE, CUSTOMER_NUM, CUSTOMER_NAME, NUM_ORDERED, QUOTED_PRICE) )
Henry Books Answer each of the following questions using the Henry Books data shown in Figures 1-4 through 1-7 in Chapter 1. No computer work is required. 1.
Ray Henry is considering expanding the activities at his book stores to include movies. He has some ideas for how he wants to do this and he needs you to help with database design activities to address these ideas. In particular, he would like you to design a database for him. He is interested in movies and wants to store information about movies, stars, and directors in a database. He needs to be able to satisfy the following requirements: a.
For each director, list his or her number, name, the year he or she was born, and the year of death if he or she is deceased.
b.
For each movie, list its number, title, the year the movie was made, and its type.
c.
For each movie, list its number, title, the number and name of its director, the critics’ rating, the MPAA rating, the number of awards for which the movie was nominated, and the number of awards the movie won.
d.
For each movie star, list his or her number, name, birthplace, the year he or she was born, and the year of death if he or she is deceased.
e.
For each movie, list its number and title, along with the number and name of all the stars who appear in it.
f.
For each movie star, list his or her number and name, along with the number and name of all the movies in which he or she stars.
List the tables, columns, and relationships. In addition, represent your design with an E-R diagram. 2.
Determine the functional dependencies that exist in the following table, and then convert this table to an equivalent collection of tables that are in third normal form. BOOK (BOOK_CODE, TITLE, TYPE, PRICE (AUTHOR_NUM, AUTHOR_LAST, AUTHOR_FIRST) )
3.
Determine the functional dependencies that exist in the following table, and then convert this table to an equivalent collection of tables that are in third normal form. BOOK (BOOK_CODE, TITLE, TYPE, PRICE, PUB_CODE, PUBLISHER_NAME, CITY)
Alexamara Marina Group Answer each of the following questions using the Alexamara Marina Group data shown in Figures 1-8 through 1-12 in Chapter 1. No computer work is required. 1.
Design a database that can satisfy the following requirements: a.
Chapter 2
For each marina, list the number, name, address, city, state, and zip code.
b.
For each boat owner, list the number, last name, first name, address, city, state, and zip code.
c.
For each marina, list all the slips in the marina. For each slip, list the length of the slip, annual rental fee, name and type of the boat occupying the slip, and boat owner’s number, last name, and first name.
d.
For each possible service category, list the category number and description. In addition, for each service request in a category, list the marina number and slip number for the boat receiving the service, estimated hours for the service, hours already spent on the service, and next date that is scheduled for the particular service.
e.
For each service request, list the marina number, slip number, category description, description of the particular service, and a description of the current status of the service.
59
List the tables, columns, and relationships. In addition, represent your design with an E-R diagram. 2.
Determine the functional dependencies that exist in the following table, and then convert this table to an equivalent collection of tables that are in third normal form. MARINA (MARINA_NUM, NAME, (SLIP_NUM, LENGTH, RENTAL_FEE, BOAT_NAME) )
3.
Determine the functional dependencies that exist in the following table, and then convert this table to an equivalent collection of tables that are in third normal form. MARINA_SLIP (SLIP_ID, MARINA_NUM, SLIP_NUM, LENGTH, RENTAL_FEE, BOAT_NAME, BOAT_TYPE, OWNER_NUM, LAST_NAME, FIRST_NAME)
Database Design Fundamentals
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CHAPTER
3
C REATI N G TABLES LEARNING
OBJECTIVES
Objectives ●
Create and run SQL commands
●
Create tables
●
Identify and use data types to define columns in tables
●
Understand and use nulls
●
Add rows to tables
●
View table data
●
Correct errors in a table
●
Save SQL commands to a file
●
Describe a table’s layout using SQL
INTRODUCTION You already might be an experienced user of a database management system (DBMS). You might find a DBMS at your school’s library, at a site on the Internet, or in any other place where you retrieve data using a computer. In this chapter, you will begin your study of Structured Query Language (SQL), which is one of the most popular and widely used languages for retrieving and manipulating database data.
In the mid-1970s, SQL was developed as the data manipulation language for IBM’s prototype relational model DBMS, System R, under the name SEQUEL at IBM’s San Jose research facilities. In 1980, the language was renamed SQL (but still pronounced “sequel” although the equally popular pronunciation of “S-Q-L” [“ess-cue-ell”] is used in this text) to avoid confusion with an unrelated hardware product named SEQUEL. Most DBMSs use a version of SQL as their data manipulation language.
In this chapter, you will learn the basics of working in SQL. You will learn how to create tables and assign data types to columns. You also will learn about a special type of value, called a null value, and 62
learn how to manage these values in tables. You will learn how to insert data into your tables after you create them. Finally, you will learn how to describe a table’s layout using SQL.
CREATING AND RUNNING SQL COMMANDS You accomplish tasks in SQL by creating and running commands. In order to do so, you need to use a DBMS that supports SQL. This text uses Oracle as the DBMS in which to create and run the commands. The text also indicates differences you will find if you are using Microsoft Access or Microsoft SQL Server 2005. (If you are using MySQL, contact Cengage Learning for the latest edition of A Guide to MySQL, by Pratt and Last.) Although the version of Oracle used in this text is the Oracle Database 10g Express Edition, the commands used in this text will work the same in any other version of Oracle. You use the Oracle Database Express Edition by downloading it from the Oracle Web site, installing it, and then starting it using the Microsoft Internet Explorer Web browser.
Starting the Oracle Database Express Edition After installing the Oracle Database Express Edition, you start it by clicking the Start button, pointing to All Programs, clicking Oracle Database 10g Express Edition, and then clicking Go To Database Home Page. Internet Explorer will start and load the home page, which requests your username and password. (Ask your instructor which username and password to use, or use the one you specified when you installed the software. If a different Web browser starts, ask your instructor for help. Other Web browsers might not fully support the examples used in this text.) After entering this information, click the Login button. Figure 3-1 shows the Oracle Database Express Edition home page. You click the icons on the home page to access the various tools. In this text, you will use the SQL tool. The other tools let you administer a database, work with database objects, and run different database utilities. (These features are beyond the scope of this book.)
Chapter 3
Currently logged on as SYSTEM user (administrator)
Help button Icons for available tools
63 Logout button
Links to additional resources SQL icon Usage monitor
FIGURE 3-1
Oracle Database Express Edition home page
NOTE The figure shows that the SYSTEM (administrator) user is currently logged on.Your instructor might assign you another username to use as your login, in which case, this name will appear on your screen.
There are two ways to use the tools in the Oracle Database Express Edition. You can click the arrow for the icon to display a menu and then select an option from the menu. Figure 3-2 shows the result of clicking the arrow for the SQL icon and then pointing to the SQL Scripts option on the SQL menu. A submenu of commands for working with SQL scripts appears. To create a script using this approach, for example, you would click the arrow, point to SQL Scripts, and then click Create. (You will learn more about scripts later in this chapter.)
Creating Tables
SQL icon arrow
64
SQL Scripts option
SQL Scripts submenu
FIGURE 3-2
SQL Scripts submenu
You also can click the icon to display the options as icons instead of as submenus. For example, when you click the SQL icon on the home page, you will see the SQL page shown in Figure 3-3. Clicking an arrow on an icon displays a submenu. The figure shows the results of clicking the arrow for the SQL Scripts icon, which displays the SQL Scripts submenu. To create a script using this approach, click the SQL icon on the home page, click the arrow for the SQL Scripts icon, and then click Create. The approach you choose is a matter of personal preference.
SQL Scripts arrow
SQL Scripts submenu
FIGURE 3-3 Chapter 3
SQL Scripts submenu on the SQL page
Entering Commands You enter commands on the SQL Commands page. To access the SQL Commands page, click the arrow for the SQL icon, and then point to SQL Commands as shown in Figure 3-4. 65 SQL icon arrow
SQL menu
SQL Commands submenu
FIGURE 3-4
Starting a new SQL command
After clicking the Enter Command option on the SQL Commands submenu, you will see the SQL Commands page shown in Figure 3-5. You enter the command in the upper portion of this page, called the SQL editor pane, and then click the Run button to execute the command and display its results in the lower portion of the page, called the Results pane.
Breadcrumbs
SQL editor pane
Run button
Results pane
FIGURE 3-5
SQL Commands page
Creating Tables
66
After clicking the Run button, the results will appear in the Results pane. Notice the Home> SQL> SQL Commands reference at the top of the SQL editor pane. This reference is called a breadcrumb. You can click the pages in the breadcrumb to move back one or more pages. For example, to return to the home page, click Home in the breadcrumb.
C R E A T I N G A TA B L E Before you begin adding data to a table, you must describe the layout of the table to the DBMS.
EXAMPLE 1 Describe the layout of the REP table to the DBMS.
You use the CREATE TABLE command to describe the layout of a table. The word TABLE is followed by the name of the table to be created and then by the names and data types of the columns that the table contains. The data type indicates the type of data that the column can contain (for example, characters, numbers, or dates) as well as the maximum number of characters or digits that the column can store. The restrictions placed on table and column names are as follows: 1. 2. 3. 4.
The The The The
names names names names
cannot exceed 30 characters. must start with a letter. can contain letters, numbers, and underscores (_). cannot contain spaces.
The SQL command that creates the REP table is shown in Figure 3-6.
Creates a table named REP
FIGURE 3-6
CREATE TABLE REP (REP_NUM CHAR(2) PRIMARY KEY, LAST_NAME CHAR(15), FIRST_NAME CHAR(15), STREET CHAR(15), CITY CHAR(15), STATE CHAR(2), ZIP CHAR(5), COMMISSION DECIMAL(7,2), RATE DECIMAL(3,2) );
Command continues over several lines
Semicolon indicates end of command
CREATE TABLE command for the REP table
This CREATE TABLE command, which uses the data definition features of SQL, describes a table named REP. The table contains nine columns: REP_NUM, LAST_NAME, FIRST_NAME, STREET, CITY, STATE, ZIP, COMMISSION, and RATE. The REP_NUM column can store two characters and is the table’s primary key. The LAST_NAME column can store 15 characters, and the STATE column can store two characters. The COMMISSION column can store only numbers, and those numbers are limited to seven digits, including two decimal places. Similarly, the RATE column can store three numbers, including two Chapter 3
decimal places. You can think of the SQL command shown in Figure 3-6 as creating an empty table with column headings for each column name. In SQL, commands are free format; that is, no rule says that a particular word must begin in a particular position on the line. For example, you could have written the CREATE TABLE command shown in Figure 3-6 as follows:
67
CREATE TABLE REP (REP_NUM CHAR(2) PRIMARY KEY, LAST_NAME CHAR(15), FIRST_NAME CHAR(15), STREET CHAR(15), CITY CHAR(15), STATE CHAR(2), ZIP CHAR(5), COMMISSION DECIMAL(7,2), RATE DECIMAL(3,2) );
The manner in which the CREATE TABLE command shown in Figure 3-6 was written makes the command more readable. This text will strive for such readability when writing SQL commands.
NOTE SQL is not case sensitive; you can type commands using uppercase or lowercase letters. There is one exception to this rule, however. When you are inserting character values into a table, you must use the correct case.
To create the REP table in Oracle, click in the SQL editor pane, type the CREATE TABLE command shown in Figure 3-7, and then click the Run button on the right side of the SQL editor pane to execute the command and create the table. Figure 3-7 also shows the message that appears in the Results pane after running the command, which indicates that the table was created.
CREATE TABLE command
Message indicates successful table creation
FIGURE 3-7
Running the CREATE TABLE command for the REP table Creating Tables
ACCESS USER NOTE
68
Microsoft Office Access is a DBMS that lets you work in a graphical user interface, but you can also use it to run SQL commands. To run SQL commands in Access, you must first create a new query. In Access 2007, open the database, click the Create tab on the Ribbon, click the Query Design button in the Other group, close the Show Table dialog box, and then click the SQL View button in the Results group. (In Access 2003, open the database, and then create a new query in Design view. Close the Show Table dialog box, and then click the SQL View button on the Query Design toolbar.) In the Query window, type the SQL command, and then click the Run button on the Query Design toolbar (Access 2003) or the Run button in the Results group on the Design tab (Access 2007) to execute the command. Figure 3-8 shows the command to create the REP table using Access 2007. When you click the Run button, Access will create the table shown in the CREATE TABLE command. (Unlike Oracle, Access doesn’t display a message indicating the result was successful.)
Run button
CREATE TABLE command
FIGURE 3-8
Using Access SQL view to create a table
Unlike Oracle, Access does not support the DECIMAL data type. To create numbers with decimals, you must use either the CURRENCY or NUMBER data type. Use the CURRENCY data type for fields that will contain currency values; use the NUMBER data type for all other numeric fields. In Access, it is common to create a table using Table Design view and then to add records to the table using Datasheet view.You still can run SQL commands when you create tables using Design view and enter data into them using Datasheet view.
Chapter 3
SQL SERVER USER NOTE Microsoft SQL Server 2005 is a DBMS designed for use in client-server applications. You can run Microsoft SQL Server 2005 from your own computer through a set of client database tools called SQL Server Management Studio. Management Studio includes a Query Editor window that you can use to run SQL commands. If you are using Management Studio and connecting to a database on your local computer, accept the default values for Server Type, Server Name, and Authentication, and then click the Connect button in the Connect to Server dialog box. When Management Studio appears, double-click Databases, open the database on which you want to run SQL commands, and click the New Query button on the toolbar. Type the SQL command in the Query Editor window that opens, and then click the Execute button on the toolbar to execute the command. The command shown in Figure 3-9 creates the REP table and displays a message in the Messages pane to indicate that the command completed successfully.
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Execute button
CREATE TABLE command
Message indicates successful table creation
FIGURE 3-9
Using Microsoft SQL Server 2005 to create a table
Correcting Errors in SQL Commands Suppose that you executed the REP table using the CREATE TABLE command shown in Figure 3-10, which contains several mistakes. Instead of displaying a message that the table was created successfully, Oracle displays an error message about a problem that it encountered. In reviewing the command, you see that CHAR is misspelled on line 4, line 5 is missing a comma, the CITY column was omitted, and line 7 should be deleted. If you run an SQL command and Oracle displays an error, you can use the mouse and the arrow keys on the keyboard to position the insertion point in the correct position so you can correct these errors using the same techniques that you might use in a word processor. For example, you can use the pointer to select the word CHR on line 4 and type CHAR. Then you can use the pointer to move the insertion point to the end of line 5 so you can type the missing comma, and then press Enter to insert the missing information to create the CITY column. You can use the pointer to select the contents of line 7 and then press Delete to remove it. After making these changes, you can click the Run button to execute the command again. If the command contains additional errors, you’ll see an error message again. If the command is correct, you’ll see the message that the table was created. Creating Tables
CHR should be CHAR
70 Missing comma
CITY column missing
XXX column should not be included
Error message
FIGURE 3-10
CREATE TABLE command with errors
Dropping a Table After creating a table, you might notice that you added a column that you do not need or that you assigned the wrong data type or size to a column. Another way of correcting errors in a table is to delete (drop) the table and start over. For example, suppose you wrote a CREATE TABLE command that contained a column named LST instead of LAST or defined a column as CHAR(5) instead of CHAR(15). Suppose you do not discover the error and you execute the command, creating a table with these problems. In this case, you can delete the entire table using the DROP TABLE command and then re-create the table using the correct CREATE TABLE command. To drop a table, execute the DROP TABLE command, followed by the name of the table you want to delete and a semicolon. To delete the REP table, for example, you would enter the following command and then click the Run button: DROP TABLE REP;
Dropping a table also deletes any data that you entered into the table. It is a good idea to check your CREATE TABLE commands carefully before executing them and to correct any problems before adding data. Later in this text, you will learn how to change a table’s structure without having to delete the entire table.
Q&A Question: How can I correct a mistake that I made when I created a table? Answer: Later in the text, you will see how to alter a table to make any necessary corrections. For now, the easiest way is to drop the table using the DROP TABLE command and then to execute the correct CREATE TABLE command.
Chapter 3
USING DATA TYPES For each column in a table, you must specify the data type to use to store the type of data that the column will contain. Figure 3-11 describes some common data types used in databases.
Data type
Description
CHAR(n)
Stores a character string n characters long. You use the CHAR data type for columns that contain letters and special characters and for columns containing numbers that will not be used in any calculations. Because neither sales rep numbers nor customer numbers will be used in any calculations, for example, the REP_NUM and CUSTOMER_NUM columns are both assigned the CHAR data type.
VARCHAR(n)
An alternative to CHAR that stores a character string up to n characters long. Unlike CHAR, only the actual character string is stored. If a character string 20 characters long is stored in a CHAR(30) column, for example, it will occupy 30 characters (20 characters plus 10 blank spaces). If it is stored in a VARCHAR(30) column, it will only occupy 20 spaces. In general, tables that use VARCHAR instead of CHAR occupy less space, but the DBMS does not process them as rapidly during queries and updates. However, both are legitimate choices. This text uses CHAR, but VARCHAR would work equally well.
DATE
Stores date data. The specific format in which dates are stored varies from one SQL implementation to another. In Oracle, dates are enclosed in single quotation marks and have the format DD-MON-YYYY (for example, '15-OCT-2010' is October 15, 2010). In Access, dates are enclosed in number signs and are entered using the format MM/DD/YYYY (for example, #10/15/2010# is October 15, 2010). In SQL Server, use the DATETIME data type to store dates.
DECIMAL(p,q)
Stores a decimal number p digits long with q of these digits being decimal places to the right of the decimal point. For example, the data type DECIMAL(5,2) represents a number with three places to the left and two places to the right of the decimal (for example, 100.00). You can use the contents of DECIMAL columns in calculations. You also can use the NUMBER (p,q) data type in both Oracle and SQL Server to store a decimal number. Access does not support the DECIMAL data type; use the CURRENCY or NUMBER data type instead.
INT
Stores integers, which are numbers without a decimal part. The valid range is -2147483648 to 2147483647. You can use the contents of INT columns in calculations. If you follow the word INT with AUTO_INCREMENT, you create a column for which SQL will automatically generate a new sequence number each time you add a new row. This would be the appropriate choice, for example, when you want the DBMS to generate a value for a primary key.
SMALLINT
Stores integers, but uses less space than the INT data type. The valid range is -32768 to 32767. SMALLINT is a better choice than INT when you are certain that the column will store numbers within the indicated range. You can use the contents of SMALLINT columns in calculations.
FIGURE 3-11
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Commonly used data types
Creating Tables
USING NULLS
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Occasionally, when you enter a new row into a table or modify an existing row, the values for one or more columns are unknown or unavailable. For example, you can add a customer’s name and address to a table even though the customer does not have an assigned sales rep or an established credit limit. In other cases, some values might never be known—perhaps there is a customer that does not have a sales rep. In SQL, you handle this situation by using a special value to represent cases in which an actual value is unknown, unavailable, or not applicable. This special value is called a null data value, or simply a null. When creating a table, you can specify whether to allow nulls in the individual columns.
Q&A Question: Should a user be allowed to enter null values for the primary key? Answer: No. The primary key is supposed to uniquely identify a given row, and this would be impossible if nulls were allowed. For example, if you stored two customer records without values in the primary key column, you would have no way to tell them apart.
In SQL, you use the NOT NULL clause in a CREATE TABLE command to indicate columns that cannot contain null values. The default is to allow nulls; columns for which you do not specify NOT NULL can accept null values. For example, suppose that the LAST_NAME and FIRST_NAME columns in the REP table cannot accept null values, but all other columns in the REP table can. The following CREATE TABLE command accomplishes this goal: CREATE TABLE REP (REP_NUM CHAR(2) PRIMARY KEY, LAST_NAME CHAR(15) NOT NULL, FIRST_NAME CHAR(15) NOT NULL, STREET CHAR(15), CITY CHAR(15), STATE CHAR(2), ZIP CHAR(5), COMMISSION DECIMAL(7,2), RATE DECIMAL(3,2) );
If you created the REP table with this CREATE TABLE command, the DBMS would reject any attempt to store a null value in either the LAST_NAME or FIRST_NAME column. The system would accept an attempt to store a null value in the STREET column, however, because the STREET column can accept null values. Because the primary key column cannot accept null values, you do not need to specify the REP_NUM column as NOT NULL.
A D D I N G R OW S T O A TA B L E After you have created a table in a database, you can load data into the table by using the INSERT command.
Chapter 3
The INSERT Command The INSERT command adds rows to a table. You type INSERT INTO followed by the name of the table into which you are adding data. Then you type the word VALUES followed by the specific values to be inserted in parentheses. When adding rows to character columns, make sure you enclose the values in single quotation marks (for example, 'Kaiser'). You also must enter the values in the appropriate case, because character data is stored exactly as you enter it.
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NOTE You must enclose values in single quotation marks for any column whose type is character (CHAR), even when the data contains numbers. Because the ZIP column in the REP table has a CHAR data type, for example, you must enclose zip codes in single quotation marks, even though they are numbers. If you need to enter an apostrophe (single quotation mark) into a column, you type two single quotation marks. For example, to enter the name O’Toole in the LAST_NAME column, you would type 'O'Toole' as the value in the INSERT command.
EXAMPLE 2 Add sales rep 20 to the REP table.
The command for this example is shown in Figure 3-12. Note that the character strings ('20','Kaiser','Valerie', and so on) are enclosed in single quotation marks. When you execute the command, the record is added to the REP table.
Character values enclosed in single quotation marks
One row was inserted
FIGURE 3-12
INSERT command for the first record in the REP table
NOTE Make sure that you type the values in the same case as those shown in the figures to avoid problems later when retrieving data from the database.
Creating Tables
EXAMPLE 3 Add sales reps 35 and 65 to the REP table. 74
You could enter and execute new INSERT commands to add the new rows to the table. However, an easier and faster way to add these new rows to the table is to use the mouse and the keyboard to modify the previous INSERT command and execute it to add the record for the second sales rep, as shown in Figure 3-13.
FIGURE 3-13
INSERT command to add the second row to the REP table
You can modify and execute the INSERT command again for the third sales rep, as shown in Figure 3-14, to add the third row to the table.
FIGURE 3-14
Chapter 3
INSERT command to add the third row to the REP table
Inserting a Row that Contains Nulls To enter a null value into a table, you use a special form of the INSERT command in which you identify the names of the columns that will accept non-null values, and then list only these non-null values after the VALUES command, as shown in Example 4.
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EXAMPLE 4 Add sales rep 85 to the REP table. Her name is Tina Webb. All columns except REP_NUM, LAST_NAME, and FIRST_NAME are null.
In this case, you do not enter a value of null; you enter only the non-null values. To do so, you must indicate precisely which values you are entering by listing the corresponding columns as shown in Figure 3-15. The command shown in Figure 3-15 indicates that you are entering data in only the REP_NUM, LAST_NAME, and FIRST_NAME columns and that you are not entering values in any other columns; the other columns will contain null values.
Only values for the listed columns will be inserted
FIGURE 3-15
Inserting a row in the REP table that contains null values
V I EW I N G TA B L E D A T A To view the data in a table, you use the SELECT command, which is described in more detail in Chapters 4 and 5.
EXAMPLE 5 Display all the rows and columns in the REP table.
Creating Tables
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You can use a simple version of the SELECT command to display all the rows and columns in a table by typing the word SELECT, followed by an asterisk, followed by the word FROM and the name of the table containing the data you want to view. Just as with other SQL commands, the command ends with a semicolon. In Oracle, you type the command shown in Figure 3-16, and then click the Run button to display the results.
Command to display all rows and all columns
FIGURE 3-16
Chapter 3
Using a SELECT command to view table data
ACCESS USER NOTE In Access, type the query shown in Figure 3-17 in SQL view to display all the rows and columns in a table.
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SELECT command
FIGURE 3-17
Using a SELECT command to view table data in Access
To run the query, click the Run button. Access will display the query results in Datasheet view, as shown in Figure 3-18. If the data does not fit on the screen, you can adjust the columns to best fit the data they contain by double-clicking the right edge of each column heading. You can use the scroll bars when necessary to view data that has scrolled off the screen.
FIGURE 3-18
Query results in Access Datasheet view
Creating Tables
SQL SERVER USER NOTE In SQL Server, type the query shown in Figure 3-19 in the Query Editor window.
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SELECT command
FIGURE 3-19
Using a SELECT command to view table data in SQL Server
To run the query, click the Execute button. SQL Server will display the query results in the Results pane, as shown in Figure 3-20. If the data does not fit on the screen, you can adjust the columns to best fit the data they contain by dragging the right edge of each column heading to make it narrower or wider.
FIGURE 3-20
Query results in SQL Server
C O R R E C T I N G E R RO RS I N A TA B L E After executing a SELECT command to view a table’s data, you might find that you need to change the value in a column. You can use the UPDATE command shown in Figure 3-21 to change a value in a table. The UPDATE command shown in Figure 3-21 changes the last name in the row on which the sales rep number is 85 to Perry.
New value Change only affects the row(s) on which the rep number is 85
Message indicates the number of updated rows
FIGURE 3-21
Chapter 3
Using an UPDATE command to change a value
The SELECT command shown in Figure 3-22 displays the results of the UPDATE command shown in Figure 3-21, in which the last name for rep number 85 is Perry. 79
FIGURE 3-22
Last name changed for sales rep number 85
When you need to delete a row from a table, you can use the DELETE command. The DELETE command shown in Figure 3-23 deletes any row on which the sales rep number is 85.
Only row(s) on which the rep number is 85 will be deleted
FIGURE 3-23
Using a DELETE command to delete a row
Creating Tables
The SELECT command shown in Figure 3-24 displays the updated data.
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FIGURE 3-24
Sales rep number 85 deleted from REP table
Q&A Question: How do I correct errors in my data? Answer: The method you use to correct an error depends on the type of error you need to correct. If you added a row that should not be in the table, use a DELETE command to remove it. If you forgot to add a row, you can use an INSERT command to add it. If you added a row that contains incorrect data, you can use an UPDATE command to make the necessary corrections. Alternatively, you could use a DELETE command to remove the row containing the error and then use an INSERT command to insert the correct row.
SAVING SQL COMMANDS Oracle lets you save a command so you can use it again without retyping it. In many DBMSs, you save commands in a script file, or simply a script, which is a text file with the .sql filename extension. When you use Oracle to create a script, Oracle stores the script in a special location called the script repository. If you want to save a script on the local file system, such as on a hard drive or USB drive, you can do so by downloading the script. When you need to use a script that is stored on the local file system, but is not currently stored in Oracle, you can upload the script so you can use it in Oracle. The following steps describe how to create and use scripts in the Oracle Database Express Edition. If you are using a different version of Oracle or another DBMS, use Help or consult the system documentation to determine how to accomplish the same tasks. To create a script: 1. 2. 3. Chapter 3
Load the Oracle Database Express Edition home page and log in. Click the SQL icon arrow, point to SQL Scripts, and then click Create. The Script Editor page opens. In the Script Name text box, type a name for the script.
4.
5.
Click in the text box on the page to activate it, and then type the command or commands to save in the script. When necessary, click the Run button to execute the commands saved in the script. When you are finished, click the Save button. You return to the SQL Scripts page and the script you created appears as an icon on the page.
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To view, edit, or run an existing script: 1. 2. 3.
4.
Load the Oracle Database Express Edition home page and log in. Click the SQL icon arrow, point to SQL Scripts, and then click View. Click the icon for the desired script. The script appears on the Script Editor page. You can use this page to view the content of the script or to make changes to it by editing the commands it contains. If you edit a script, click the Save button to save your changes. To run a script, click the Run button. The Run Script page loads and asks you to confirm running the script. Click the Run button on the Run Script page. The Manage Script Results page opens and displays the script name and an icon in the View Results column. To see the results of the command stored in the script, click the icon in the View Results column.
When you are finished using a script or no longer need to store it, you can delete it. To delete a script: 1. 2. 3.
Follow the previous instructions to view the script. Click the Delete button on the Script Editor page. Click the OK button to confirm the deletion.
To download a script from the script repository so you can save it as a file: 1. 2. 3. 4.
Follow the previous instructions to view the script. Click the Download button. The File Download dialog box opens. In the dialog box, click the Save button, and then save the file to the desired location. Click the Close button to close the Download complete dialog box.
To upload a script to the script repository: 1. 2. 3.
4.
Load the Oracle Database Express Edition home page and log in. Click the SQL icon arrow, point to SQL Scripts, and then click Upload. The Upload Script page opens. Click the Browse button. The Choose file dialog box opens. Navigate to and select the script file to upload. Click the Open button. (If you want to upload the script with a different filename, type the new name in the Script Name text box.) On the Upload Script page, click the Upload button. An icon for the script appears on the SQL Scripts page.
Creating Tables
ACCESS USER NOTE
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Access does not use script files, but you can save an SQL command as a query object in the database. Open the database, create the query in SQL view, click the Save button on the Quick Access toolbar (or on the menu bar), and then save the query with the desired object name. To run the query without first viewing the SQL command, right-click the query in the Navigation Pane (or in the Database window), and then click Open on the shortcut menu. The query results will appear in Datasheet view. If you want to view the SQL command before running the query, right-click the query in the Navigation Pane, and then click Design View on the shortcut menu. To run the SQL command, click the Run button.
SQL SERVER USER NOTE SQL Server can store scripts in any folder on your local system. All scripts created in SQL Server are text files with the .sql extension. To create a script file in SQL Server: 1. Load SQL Server Management Studio, and then click the Connect button in the Connect to Server dialog box. 2. Open the appropriate database, and then click the New Query button. 3. Type the command or commands to save in the script. When necessary, click the Execute button to execute the commands saved in the script. 4. When you are finished, click the Save button, navigate to the location in which to save the script, and then enter a name for the script. To view, edit, or run an existing script: 1. Load SQL Server Management Studio, and then click the Connect button in the Connect to Server dialog box. 2. Open the appropriate database, and then click the New Query button. 3. Click the Open File button on the toolbar. 4. Navigate to the folder containing the script file, select the script file, and then click the Open button in the Open File dialog box. (If the Connect to Database Engine dialog box opens, click the Connect button.) The script appears in the Query Editor window. You can view the contents of the script to make changes to it by editing the commands. If you edit a script, click the Save button to save your changes. 5. To run a script, click the Execute button.
CREATING THE REMAINING DATABASE TA B L E S To create the remaining tables in the Premiere Products database, you need to execute the appropriate CREATE TABLE and INSERT commands. You should save these commands as scripts so you can re-create your database, if necessary, by running the scripts.
NOTE Your instructor might give you the script files to use to create the tables in the Premiere Products, Henry Books, and Alexamara Marina Group databases and to insert data into them.
Chapter 3
Figure 3-25 shows the CREATE TABLE command for the CUSTOMER table. Notice that the CUSTOMER_NAME column is specified as NOT NULL. Additionally, the CUSTOMER_NUM column is the table’s primary key, indicating that the CUSTOMER_NUM column is the unique identifier of rows in the table. With this column designated as the primary key, the DBMS will reject any attempt to store a customer number that already exists in the table.
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CUSTOMER_NUM is the primary key CREATE TABLE CUSTOMER (CUSTOMER_NUM CHAR(3) PRIMARY KEY, CUSTOMER_NAME CHAR(35) NOT NULL, CUSTOMER_NAME STREET CHAR(15), cannot be null CITY CHAR(15), STATE CHAR(2), Semicolon indicates ZIP CHAR(5), end of command BALANCE DECIMAL(8,2), CREDIT_LIMIT DECIMAL(8,2), REP_NUM CHAR(2) );
FIGURE 3-25
CREATE TABLE command for the CUSTOMER table
After creating the CUSTOMER table, you can create another script file containing the INSERT commands to add the customer rows to the table. When a script file contains more than one command, each command must end with a semicolon. Figure 3-26 shows the INSERT commands to add rows to the CUSTOMER table. As noted previously, to enter an apostrophe (single quotation mark) in the value for a field, type two single quotation marks, as illustrated in the name in the first INSERT command (Al’s Appliance and Sport) in Figure 3-26.
Creating Tables
Type two single quotation marks to insert an apostrophe in a value
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Data for
INSERT INTO CUSTOMER first row VALUES ('148','Al's Appliance and Sport','2837 Greenway','Fillmore','FL','33336',6550.00,7500.00,'20'); INSERT INTO CUSTOMER VALUES Data for ('282','Brookings Direct','3827 Devon','Grove','FL','33321',431.50,10000.00,'35'); second row INSERT INTO CUSTOMER VALUES ('356','Ferguson's','382 Wildwood','Northfield','FL','33146',5785.00,7500.00,'65'); INSERT INTO CUSTOMER VALUES ('408','The Everything Shop','1828 Raven','Crystal','FL','33503',5285.25,5000.00,'35'); INSERT INTO CUSTOMER VALUES ('462','Bargains Galore','3829 Central','Grove','FL','33321',3412.00,10000.00,'65'); INSERT INTO CUSTOMER VALUES Each command ('524','Kline's','838 Ridgeland','Fillmore','FL','33336',12762.00,15000.00,'20'); ends with a INSERT INTO CUSTOMER semicolon VALUES ('608','Johnson's Department Store','372 Oxford','Sheldon','FL','33553',2106.00,10000.00,'65'); INSERT INTO CUSTOMER VALUES ('687','Lee's Sport and Appliance','282 Evergreen','Altonville','FL','32543',2851.00,5000.00,'35'); INSERT INTO CUSTOMER VALUES ('725','Deerfield's Four Seasons','282 Columbia','Sheldon','FL','33553',248.00,7500.00,'35'); INSERT INTO CUSTOMER VALUES ('842','All Season','28 Lakeview','Grove','FL','33321',8221.00,7500.00,'20');
FIGURE 3-26
INSERT commands for the CUSTOMER table
Figures 3-27 through 3-32 show the scripts for the CREATE TABLE and INSERT commands for creating and inserting data into the ORDERS, PART, and ORDER_LINE tables in the Premiere Products database. Figure 3-27 contains the CREATE TABLE command for the ORDERS table. CREATE TABLE ORDERS (ORDER_NUM CHAR(5) PRIMARY KEY, ORDER_DATE DATE, CUSTOMER_NUM CHAR(3) );
FIGURE 3-27
ORDER_NUM is the primary key
CREATE TABLE command for the ORDERS table
Figure 3-28 contains the INSERT commands to load data into the ORDERS table. Notice the way that dates are entered.
Chapter 3
INSERT INTO ORDERS VALUES ('21608','20-OCT-2010','148'); INSERT INTO ORDERS VALUES ('21610','20-OCT-2010','356'); INSERT INTO ORDERS VALUES ('21613','21-OCT-2010','408'); INSERT INTO ORDERS VALUES ('21614','21-OCT-2010','282'); INSERT INTO ORDERS VALUES ('21617','23-OCT-2010','608'); INSERT INTO ORDERS VALUES ('21619','23-OCT-2010','148'); INSERT INTO ORDERS VALUES ('21623','23-OCT-2010','608');
FIGURE 3-28
Format used to enter a date in Oracle
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INSERT commands for the ORDERS table
Figure 3-29 contains the CREATE TABLE command for the PART table. CREATE TABLE PART (PART_NUM CHAR(4) PRIMARY KEY, DESCRIPTION CHAR(15), ON_HAND DECIMAL(4,0), CLASS CHAR(2), WAREHOUSE CHAR(1), PRICE DECIMAL(6,2) );
FIGURE 3-29
PART_NUM is the primary key
CREATE TABLE command for the PART table
Creating Tables
Figure 3-30 contains the INSERT commands to load data into the PART table.
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INSERT INTO PART VALUES ('AT94','Iron',50,'HW','3',24.95); INSERT INTO PART VALUES ('BV06','Home Gym',45,'SG','2',794.95); INSERT INTO PART VALUES ('CD52','Microwave Oven',32,'AP','1',165.00); INSERT INTO PART VALUES ('DL71','Cordless Drill',21,'HW','3',129.95); INSERT INTO PART VALUES ('DR93','Gas Range',8,'AP','2',495.00); INSERT INTO PART VALUES ('DW11','Washer',12,'AP','3',399.99); INSERT INTO PART VALUES ('FD21','Stand Mixer',22,'HW','3',159.95); INSERT INTO PART VALUES ('KL62','Dryer',12,'AP','1',349.95); INSERT INTO PART VALUES ('KT03','Dishwasher',8,'AP','3',595.00); INSERT INTO PART VALUES ('KV29','Treadmill',9,'SG','2',1390.00);
FIGURE 3-30
INSERT commands for the PART table
Figure 3-31 contains the CREATE TABLE command for the ORDER_LINE table. Notice the way that the primary key is defined when it consists of more than one column. CREATE TABLE ORDER_LINE (ORDER_NUM CHAR(5), PART_NUM CHAR(4), NUM_ORDERED DECIMAL(3,0), QUOTED_PRICE DECIMAL(6,2), PRIMARY KEY (ORDER_NUM, PART_NUM) );
The combination ORDER_NUM and PART_NUM is the primary key
FIGURE 3-31
Primary key must be entered in this format when it consists of more than one column
CREATE TABLE command for the ORDER_LINE table
Figure 3-32 contains the INSERT commands to load data into the ORDER_LINE table.
Chapter 3
INSERT INTO ORDER_LINE VALUES ('21608','AT94',11,21.95); INSERT INTO ORDER_LINE VALUES ('21610','DR93',1,495.00); INSERT INTO ORDER_LINE VALUES ('21610','DW11',1,399.99); INSERT INTO ORDER_LINE VALUES ('21613','KL62',4,329.95); INSERT INTO ORDER_LINE VALUES ('21614','KT03',2,595.00); INSERT INTO ORDER_LINE VALUES ('21617','BV06',2,794.95); INSERT INTO ORDER_LINE VALUES ('21617','CD52',4,150.00); INSERT INTO ORDER_LINE VALUES ('21619','DR93',1,495.00); INSERT INTO ORDER_LINE VALUES ('21623','KV29',2,1290.00);
FIGURE 3-32
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INSERT commands for the ORDER_LINE table
D E S C R I B I N G A TA B L E The CREATE TABLE command defines a table’s structure by listing its columns, data types, and column lengths. The CREATE TABLE command also indicates which columns cannot accept nulls. When you work with a table, you might not have access to the CREATE TABLE command that was used to create it. For example, another programmer might have created the table, or perhaps you created the table several months ago but did not save the command. You might want to examine the table’s structure to see the details about the columns in the table. Each DBMS provides a method to examine a table’s structure.
EXAMPLE 6 Describe the REP table.
In Oracle, you can use the DESCRIBE command to list all the columns in a table and their properties. Figure 3-33 shows the DESCRIBE command for the REP table. The result indicates the name of each column in the table, along with its data type and length. A value of 1 in the Primary Key column indicates the table’s primary key column. A check mark in the Nullable column indicates a column that can accept null values. (The Precision, Scale, Default, and Comment columns in the results are beyond the scope of this discussion.)
Creating Tables
DESCRIBE command and table name A check mark indicates a column can accept null values
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FIGURE 3-33
DESCRIBE command for the REP table
ACCESS USER NOTE In Access, you use the Documenter tool to describe the tables (and other objects) in a database. To start the Documenter in Access 2003, open the database, click Tools on the menu bar, point to Analyze, and then click Documenter. To start the Documenter in Access 2007, click the Database Tools tab on the Ribbon, and then click the Database Documenter button in the Analyze group. In the Documenter dialog box, click the Tables tab, select the tables that you want to describe by putting a check mark in the check box next to their names, and then click the OK button. The Object Definition window opens and displays a report containing the requested documentation. You can customize the Documenter to control the amount of detail included in the report.
SQL SERVER USER NOTE In SQL Server, you execute the sp_columns command to list all the columns in a table. The following command will list all the columns in the REP table: Exec sp_columns REP The result will indicate the name of each column in the REP table, along with its data type and length. A value of 1 in the Nullable column indicates a column that can accept null values. (The remaining columns that appear in the results are beyond the scope of this discussion.)
Chapter 3
Chapter Summary ●
Use the CREATE TABLE command to create a table by typing the table name and then listing within a single set of parentheses the columns in the table.
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Use the DROP TABLE command to delete a table and all its data from the database.
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Some commonly used data types in are INT, SMALLINT, DECIMAL, CHAR, VARCHAR, and DATE. Microsoft Access does not support DECIMAL. SQL Server uses DATETIME instead of DATE.
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A null data value (or null) is a special value that is used when the actual value for a column is unknown, unavailable, or not applicable.
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Use the NOT NULL clause in a CREATE TABLE command to identify columns that cannot accept null values.
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Use the INSERT command to insert rows into a table.
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Use the SELECT command to view the data in a table.
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Use the UPDATE command to change the value in a column.
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Use the DELETE command to delete a row from a table.
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You can save SQL commands in a script file in Oracle and SQL Server. In Microsoft Access, you save the commands as a query object in the database.
●
You can use the DESCRIBE command in Oracle to display a table’s structure and layout. In Access, you use the Documenter tool to produce a report of a table’s structure and layout. In SQL Server, execute the sp_columns command to display the structure and layout of a table.
89
Key Terms breadcrumb
null
CREATE TABLE
null data value
data type
script
DELETE
script file
DESCRIBE
script repository
DROP TABLE
SELECT
INSERT
Structured Query Language (SQL)
NOT NULL
UPDATE
Review Questions 1.
How do you create a table using SQL?
2.
How do you delete a table using SQL?
3.
What are the common data types used to define columns using SQL?
Creating Tables
4.
90
Identify the best data type to use to store the following data in Oracle, in SQL Server, and in Access: a.
The month, day, and year that an employee was hired
b.
An employee’s Social Security number
c.
The department in which an employee works
d.
An employee’s hourly pay rate
5.
Write a paragraph that explains the difference between the CHAR data type and the VARCHAR data type. Use your Web browser and your favorite search engine to find examples of when to use VARCHAR and when to use CHAR. Be sure to cite the URL(s) that provided the examples as references at the end of your document.
6.
What is a null value? How do you use SQL to identify columns that cannot accept null values?
7.
Which SQL command do you use to add a row to a table?
8.
Which SQL command do you use to view the data in a table?
9.
Which SQL command do you use to change the value in a column in a table?
10.
Which SQL command do you use to delete rows from a table?
11.
How do you display the columns in a table and their characteristics in Oracle?
Exercises To print a copy of your commands and results using Oracle, use the browser’s Print command on the File menu or click the Print button on the browser’s toolbar. To print a copy of your commands and results using Access, use the instructions provided in the chapter to save your commands as query objects. To print a command, start Word or another word processor and create a new document. Select the SQL command in Access, copy it to the Clipboard, and then paste it into the document. To copy and paste a command’s results, rightclick the datasheet selector (the box in the upper-left corner of the datasheet) to select the entire datasheet, copy it to the Clipboard, and then paste it into the document. To print a copy of your commands and results using SQL Server, start Word or another word processor and create a new document. Select the SQL command in SQL Server, copy it to the Clipboard, and then paste it into the document. To copy and paste a command’s results, rightclick the datasheet selector (the box in the upper-left corner of the datasheet) to select the entire datasheet, copy it to the Clipboard, and then paste it into the document.
Premiere Products Use SQL to complete the following exercises. 1.
Chapter 3
Create a table named SALES_REP. The table has the same structure as the REP table shown in Figure 3-6 except the LAST_NAME column should use the VARCHAR data type and the COMMISSION and RATE columns should use the NUMBER data type. Execute the command to describe the layout and characteristics of the SALES_REP table.
2.
Add the following row to the SALES_REP table: rep number: 25, last name: Lim; first name: Louis; street: 535 Vincent; city: Grove; state: FL; zip code: 33321; commission: 0.00; and rate: 0.05. Display the contents of the SALES_REP table.
3.
Delete the SALES_REP table.
4.
Run the script file for the Premiere database to create the five tables and add records to the tables. Be sure to select the file for the particular DBMS that you are using (Oracle, SQL Server, or Access). (Note: If you do not have the files for this text, ask your instructor for assistance.)
5.
Confirm that you have created the tables correctly by describing each table and comparing the results to Figures 3-6, 3-25, 3-27, 3-29, and 3-31.
6.
Confirm that you have added all data correctly by viewing the data in each table and comparing the results to Figure 2-1 in Chapter 2.
91
Henry Books Use SQL to complete the following exercises. 1.
Create a table named SALES_BRANCH. The table has the same structure as the BRANCH table shown in Figure 3-34 except the BRANCH_LOCATION column should use the VARCHAR data type and the BRANCH_NUM and NUM_EMPLOYEES columns should use the NUMBER data type. Execute the command to describe the layout and characteristics of the SALES_BRANCH table.
2.
Add the following row to the SALES_BRANCH table: branch number: 5; branch name: Henry Town Plaza; branch location: 165 Plaza; and number of employees: 3. Display the contents of the SALES_BRANCH table.
3.
Delete the SALES_BRANCH table.
4.
Run the script file for the Henry Books database to create the six tables and add records to the tables. Be sure to select the file for the particular DBMS that you are using (Oracle, SQL Server, or Access). (Note: If you do not have the files for this text, ask your instructor for assistance.)
5.
Confirm that you have created the tables correctly by describing each table and comparing the results to Figure 3-34.
6.
Confirm that you have added all data correctly by viewing the data in each table and comparing the results to Figures 1-4 through 1-7 in Chapter 1.
Creating Tables
BRANCH
92
Column
Type
Length
Decimal places
Nulls allowed?
Description
BRANCH_NUM
DECIMAL
2
0
No
Branch number (primary key)
BRANCH_NAME
CHAR
50
Branch name
BRANCH_LOCATION
CHAR
50
Branch location
NUM_EMPLOYEES
DECIMAL
2
0
Column
Type
Length
Decimal places
PUBLISHER_CODE
CHAR
3
PUBLISHER_NAME
CHAR
25
Publisher name
CITY
CHAR
20
Publisher city
Column
Type
Length
Decimal places
Nulls allowed?
Description
AUTHOR_NUM
DECIMAL
2
0
No
Author number (primary key)
AUTHOR_LAST
CHAR
12
Author last name
AUTHOR_FIRST
CHAR
10
Author first name
Column
Type
Length
BOOK_CODE
CHAR
4
TITLE
CHAR
40
Book title
PUBLISHER_CODE
CHAR
3
Publisher code
TYPE
CHAR
3
Book type
PRICE
DECIMAL
4
PAPERBACK
CHAR
1
Number of employees
PUBLISHER Nulls allowed?
Description
No
Publisher code (primary key)
AUTHOR
BOOK
FIGURE 3-34
Chapter 3
Decimal places
2
Table layouts for the Henry Books database
Nulls allowed?
Description
No
Book code (primary key)
Book price Paperback (Y, N)
WROTE Column
Type
Length
Decimal places
BOOK_CODE
CHAR
4
AUTHOR_NUM
DECIMAL
2
0
SEQUENCE
DECIMAL
1
0
Column
Type
Length
Decimal places
BOOK_CODE
CHAR
4
BRANCH_NUM
DECIMAL
2
0
ON_HAND
DECIMAL
2
0
Nulls allowed?
Description
No
Book code (primary key)
No
Author number (primary key)
93
Sequence number
INVENTORY
FIGURE 3-34
Nulls allowed?
Description
No
Book code (primary key)
No
Branch number (primary key) Units on hand
Table layouts for the Henry Books database (continued)
Creating Tables
Alexamara Marina Group Use SQL to complete the following exercises. 1.
Create a table named BOAT_SLIP. The table has the same structure as the MARINA_SLIP table shown in Figure 3-35 except the SLIP_ID, LENGTH, and RENTAL_FEE columns should use the NUMBER data type. Execute the command to describe the layout and characteristics of the BOAT_SLIP table.
2.
Add the following record to the BOAT_SLIP table: slip ID: 12; marina number: 2; slip number: 7; length: 25; rental fee: 1800; boat name: Bavant; boat type: Ray 25; and owner number: FL13. Display the contents of the BOAT_SLIP table.
3.
Delete the BOAT_SLIP table.
4.
Run the script file for the Alexamara Marina Group database to create the five tables and add records to the tables. Be sure to select the file for the particular DBMS that you are using (Oracle, SQL Server, or Access). (Note: If you do not have the files for this text, ask your instructor for assistance.)
5.
Confirm that you have created the tables correctly by describing each table and comparing the results to Figure 3-35.
6.
Confirm that you have added all data correctly by viewing the data in each table and comparing the results to Figures 1-8 through 1-12 in Chapter 1.
94
Chapter 3
MARINA Column
Type
Length
Decimal places
Nulls allowed?
Description
MARINA_NUM
CHAR
4
No
Marina number (primary key)
NAME
CHAR
20
Marina name
ADDRESS
CHAR
15
Marina street address
CITY
CHAR
15
Marina city
STATE
CHAR
2
Marina state
ZIP
CHAR
5
Marina zip code
Column
Type
Length
OWNER_NUM
CHAR
4
LAST_NAME
CHAR
50
Owner last name
FIRST_NAME
CHAR
20
Owner first name
ADDRESS
CHAR
15
Owner street address
CITY
CHAR
15
Owner city
STATE
CHAR
2
Owner state
ZIP
CHAR
5
Owner zip code
Column
Type
Length
Decimal places
Nulls allowed?
Description
SLIP_ID
DECIMAL
4
0
No
Slip ID (primary key)
MARINA_NUM
CHAR
4
Marina number
SLIP_NUM
CHAR
4
Slip number in the marina
LENGTH
DECIMAL
4
0
Length of slip (in feet)
RENTAL_FEE
DECIMAL
8
2
Annual rental fee for the slip
BOAT_NAME
CHAR
50
Name of boat currently in the slip
BOAT_TYPE
CHAR
50
Type of boat currently in the slip
OWNER_NUM
CHAR
4
Number of boat owner renting the slip
95
OWNER Decimal places
Nulls allowed?
Description
No
Owner number (primary key)
MARINA_SLIP
FIGURE 3-35
Table layouts for the Alexamara Marina Group database
Creating Tables
SERVICE_CATEGORY
96
Column
Type
Length
Decimal places
Nulls allowed?
Description
CATEGORY_NUM
DECIMAL
4
0
No
Category number (primary key)
CATEGORY_ DESCRIPTION
CHAR
255
Column
Type
Length
Decimal places
Nulls allowed?
Description
SERVICE_ID
DECIMAL
4
0
No
Service ID (primary key)
SLIP_ID
DECIMAL
4
0
Slip ID of the boat for which service is requested
CATEGORY_NUM
DECIMAL
4
0
Category number of the requested service
DESCRIPTION
CHAR
255
Description of specific service requested for boat
STATUS
CHAR
255
Description of status of service request
EST_HOURS
DECIMAL
4
2
Estimated number of hours required to complete the service
SPENT_HOURS
DECIMAL
4
2
Hours already spent on the service
NEXT_SERVICE_DATE
DATE
Category description
SERVICE_REQUEST
FIGURE 3-35
Chapter 3
Next scheduled date for work on this service (or null if no next service date is specified)
Table layouts for the Alexamara Marina Group database (continued)
CHAPTER
4
S I NGLE- TAB LE Q UER IES LEARNING
OBJECTIVES
Objectives ●
Retrieve data from a database using SQL commands
●
Use simple and compound conditions in queries
●
Use the BETWEEN, LIKE, and IN operators in queries
●
Use computed columns in queries
●
Sort data using the ORDER BY clause
●
Sort data using multiple keys and in ascending and descending order
●
Use aggregate functions in a query
●
Use subqueries
●
Group data using the GROUP BY clause
●
Select individual groups of data using the HAVING clause
●
Retrieve columns with null values
INTRODUCTION In this chapter, you will learn about the SQL SELECT command that is used to retrieve data in a database.You will examine ways to sort data and use SQL functions to count rows and calculate totals.You also will learn how to nest SELECT commands by placing one SELECT command inside another. Finally, you will learn how to group rows that have matching values in some column.
CONSTRUCTING SIMPLE QUERIES
98
One of the most important features of a DBMS is its ability to answer a wide variety of questions concerning the data in a database. When you need to find data that answers a specific question, you use a query. A query is a question represented in a way that the DBMS can understand. In SQL, you use the SELECT command to query a database. The basic form of the SELECT command is SELECT-FROM-WHERE. After you type the word SELECT, you list the columns that you want to include in the query results. This portion of the command is called the SELECT clause. Next, you type the word FROM followed by the name of the table that contains the data you need to query. This portion of the command is called the FROM clause. Finally, after the word WHERE, you list any conditions (restrictions) that apply to the data you want to retrieve. This optional portion of the command is called the WHERE clause. For example, when you need to retrieve the rows for only those customers with credit limits of $7,500, include a condition in the WHERE clause specifying that the value in the CREDIT_LIMIT column must be $7,500 (CREDIT_LIMIT = 7500). There are no special formatting rules in SQL. In this text, the FROM clause and the WHERE clause (when it is used) appear on separate lines only to make the commands more readable and understandable.
Retrieving Certain Columns and All Rows You can write a command to retrieve specified columns and all rows from a table, as illustrated in Example 1.
EXAMPLE 1 List the number, name, and balance for all customers.
Because you need to list all customers, you do not need to include a WHERE clause; you do not need to put any restrictions on the data to retrieve. You simply list the columns to be included (CUSTOMER_NUM, CUSTOMER_NAME, and BALANCE) in the SELECT clause and the name of the table (CUSTOMER) in the FROM clause. Type a semicolon to indicate the end of the command, and then click the Run button to display the results. The query and its results appear in Figure 4-1.
Chapter 4
Display list box Columns to include in the query results
99
Data will come from the CUSTOMER table
Query results
FIGURE 4-1
SELECT command to select certain columns from the CUSTOMER table
NOTE In the Oracle Database Express Edition, the number in the Display list box indicates the maximum number of rows that Oracle will display in the query results. The default value is 10. To change the value, either click the arrow and select a new value from the list or type a new value in the box. Figure 4-1 shows the Display list box after the user changed it to display 100 rows. When you run a query whose results will include more rows than the number in the Display list box, Oracle will display a message indicating this fact. If this situation occurs, increase the number in the Display list box, and then click the Run button again to display the complete query results.
NOTE If you are using Access or SQL Server to run the SQL commands shown in this text, your query results will differ slightly from the results shown in the figures. In Access, the BALANCE field has the CURRENCY data type and Access will display values in this column with two decimal places and a dollar sign. In SQL Server, values in the BALANCE field will be displayed with two decimal places and DATE field values might be displayed with a time value. Although your output might be formatted differently, the data should be the same as what you see in the figures.
Single-Table Queries
Retrieving All Columns and All Rows You can use the same type of command illustrated in Example 1 to retrieve all columns and all rows from a table. As Example 2 illustrates, however, you can use a shortcut to accomplish this task. 100
EXAMPLE 2 List the complete PART table.
Instead of including every column in the SELECT clause, you can use an asterisk (*) to indicate that you want to include all columns. The result lists all columns in the order in which you described them to the DBMS when you created the table. If you want the columns listed in a different order, type the column names in the order in which you want them to appear in the query results. In this case, assuming that the default order is appropriate, you can use the query shown in Figure 4-2 to display the complete PART table.
Asterisk indicates all columns will be included
FIGURE 4-2
SELECT command to select all columns from the PART table
Using a WHERE Clause When you need to retrieve rows that satisfy some condition, you include a WHERE clause in the SELECT command, as shown in Example 3.
Chapter 4
EXAMPLE 3 What is the name of the customer with customer number 148?
You can use a WHERE clause to restrict the query results to customer number 148, as shown in Figure 4-3. Because CUSTOMER_NUM is a character column, the value 148 is enclosed in single quotation marks. In addition, because the CUSTOMER_NUM column is the primary key of the CUSTOMER table, there can be only one customer whose number matches the number in the WHERE clause.
101
Value is enclosed in single quotation marks because CUSTOMER_NUM is a character column
FIGURE 4-3
SELECT command to find the name of customer number 148
The condition in the preceding WHERE clause is called a simple condition. A simple condition has the form column name, comparison operator, and then either another column name or a value. Figure 4-4 lists the comparison operators that you can use in SQL. Notice that there are two versions of the “not equal to” operator: < > and !=.
Comparison operator
Description
=
Equal to
<
Less than
>
Greater than
<=
Less than or equal to
>=
Greater than or equal to
<>
Not equal to
!=
Not equal to
FIGURE 4-4
Comparison operators used in SQL commands
Single-Table Queries
EXAMPLE 4 Find the number and name of each customer located in the city of Grove.
102
The only difference between this example and the previous one is that in Example 3, there could not be more than one row in the answer because the condition involved the table’s primary key. In Example 4, the condition involves a column that is not the table’s primary key. Because there is more than one customer located in the city of Grove, the results can and do contain more than one row, as shown in Figure 4-5.
Condition
FIGURE 4-5
SELECT command to find all customers located in Grove
EXAMPLE 5 Find the number, name, balance, and credit limit for all customers with balances that exceed their credit limits.
A simple condition can also compare the values stored in two columns. In Figure 4-6, the WHERE clause includes a comparison operator that selects only those rows in which the balance is greater than the credit limit.
Chapter 4
103
FIGURE 4-6
SELECT command to find all customers with balances that exceed their credit limits
Using Compound Conditions The conditions you have seen so far are called simple conditions. The following examples require compound conditions. You form a compound condition by connecting two or more simple conditions with the AND, OR, and NOT operators. When the AND operator connects simple conditions, all the simple conditions must be true in order for the compound condition to be true. When the OR operator connects the simple conditions, the compound condition will be true whenever any one of the simple conditions is true. Preceding a condition by the NOT operator reverses the truth of the original condition. For example, if the original condition is true, the new condition will be false; if the original condition is false, the new one will be true.
EXAMPLE 6 List the descriptions of all parts that are located in warehouse 3 and for which there are more than 25 units on hand.
In Example 6, you need to retrieve those parts that meet both conditions—the warehouse number is equal to 3 and the number of units on hand is greater than 25. To find the answer, you form a compound condition using the AND operator, as shown in Figure 4-7. The query examines the data in the PART table and lists the parts that are located in warehouse 3 and for which there are more than 25 units on hand. When a WHERE clause uses the AND operator to connect simple conditions, it also is called an AND condition.
Single-Table Queries
AND condition
104
FIGURE 4-7
SELECT command with an AND condition on separate lines
For readability, each of the simple conditions in the query shown in Figure 4-7 appears on a separate line. Some people prefer to put the conditions on the same line with parentheses around each simple condition, as shown in Figure 4-8. These two methods accomplish the same thing. In this text, simple conditions will appear on separate lines and without parentheses.
FIGURE 4-8
SELECT command with an AND condition on a single line
EXAMPLE 7 List the descriptions of all parts that are located in warehouse 3 or for which there are more than 25 units on hand.
In Example 7, you need to retrieve descriptions for those parts for which the warehouse number is equal to 3, or the number of units on hand is greater than 25, or both. To do this, you form a compound condition using the OR operator, as shown in Figure 4-9. When a WHERE clause uses the OR operator to connect simple conditions, it also is called an OR condition.
Chapter 4
OR condition
105
FIGURE 4-9
SELECT command with an OR condition
EXAMPLE 8 List the descriptions of all parts that are not in warehouse 3.
For Example 8, you could use a simple condition and the “not equal to” operator (WHERE WAREHOUSE < > '3'). As an alternative, you could use the EQUAL operator (=) in the condition and precede the entire condition with the NOT operator, as shown in Figure 4-10. When a WHERE clause uses the NOT operator to connect simple conditions, it also is called a NOT condition.
NOT condition
FIGURE 4-10
SELECT command with a NOT condition Single-Table Queries
You do not need to enclose the condition WAREHOUSE = '3' in parentheses, but doing so makes the command more readable.
Using the BETWEEN Operator Example 9 requires a compound condition to determine the answer. 106
EXAMPLE 9 List the number, name, and balance of all customers with balances greater than or equal to $2,000 and less than or equal to $5,000.
You can use a WHERE clause and the AND operator, as shown in Figure 4-11, to retrieve the data.
FIGURE 4-11
SELECT command with an AND condition for a single column
NOTE In SQL, numbers included in queries are entered without extra symbols, such as dollar signs and commas.
An alternative to this approach uses the BETWEEN operator, as shown in Figure 4-12. The BETWEEN operator lets you specify a range of values in a condition.
Chapter 4
107
FIGURE 4-12
SELECT command with the BETWEEN operator
The BETWEEN operator is inclusive, meaning that the query selects a value equal to either value in the condition and in the range of the values. In the clause BETWEEN 2000 and 5000, for example, values of 2,000 through 5,000 would make the condition true. You can use the BETWEEN operator in Oracle, SQL Server, and Access. The BETWEEN operator is not an essential feature of SQL; you have just seen that you can obtain the same result without it. Using the BETWEEN operator, however, does make certain SELECT commands simpler to construct.
Using Computed Columns You can perform computations using SQL queries. A computed column does not exist in the database but can be computed using data in the existing columns. Computations can involve any arithmetic operator shown in Figure 4-13.
Arithmetic operator
Description
+
Addition
–
Subtraction
*
Multiplication
/
Division
FIGURE 4-13
Arithmetic operators
EXAMPLE 10 Find the number, name, and available credit (the credit limit minus the balance) for each customer.
Single-Table Queries
There is no column in the Premiere Products database that stores a customer’s available credit, but you can compute the available credit using the CREDIT_LIMIT and BALANCE columns. To compute the available credit, you use the expression CREDIT_LIMIT BALANCE, as shown in Figure 4-14. 108
Computation
Computed column results
FIGURE 4-14
SELECT command with a computed column
The parentheses around the calculation (CREDIT_LIMIT - BALANCE) are not essential but improve readability. You also can assign a name to a computed column by following the computation with the word AS and the desired name. The command shown in Figure 4-15, for example, assigns the name AVAILABLE_CREDIT to the computed column.
Chapter 4
Computed column name
FIGURE 4-15
109
SELECT command with a named computed column
NOTE You can use names containing spaces following the word AS. In many SQL implementations, including Oracle, you do so by enclosing the name in quotation marks (for example, AS 'AVAILABLE CREDIT'). Other SQL implementations require you to enclose the name in other special characters. For example, in Access you would enclose the name in square brackets (AS [AVAILABLE CREDIT]). In SQL Server, you can use either quotation marks or square brackets.
EXAMPLE 11 Find the number, name, and available credit for each customer with at least $5,000 of available credit.
You also can use computed columns in comparisons, as shown in Figure 4-16.
Single-Table Queries
110
FIGURE 4-16
SELECT command with a computation in the condition
Using the LIKE Operator In most cases, the conditions in WHERE clauses involve exact matches, such as retrieving rows for each customer located in the city of Grove. In some cases, however, exact matches do not work. For example, you might know that the desired value contains only a certain collection of characters. In such cases, you use the LIKE operator with a wildcard symbol, as shown in Example 12. Rather than testing for equality, the LIKE operator uses one or more wildcard characters to test for a pattern match.
EXAMPLE 12 List the number, name, and complete address of each customer located on a street that contains the letters “Central.”
All you know is that the addresses you want contain a certain collection of characters (“Central”) somewhere in the STREET column, but you do not know where. In SQL for Oracle and for SQL Server, the percent sign (%) is used as a wildcard to represent any collection of characters. As shown in Figure 4-17, the condition LIKE '%Central%' retrieves information for each customer whose street contains some collection of characters, followed by the letters “Central,” followed potentially by some additional characters. Note that this query also would retrieve information for a customer whose address is “123 Centralia” because “Centralia” also contains the letters “Central.”
Chapter 4
Wildcard symbols used in condition
111 Street contains “Central”
FIGURE 4-17
SELECT command with a LIKE operator and wildcards
Another wildcard symbol in SQL is the underscore (_), which represents any individual character. For example, 'T_m' represents the letter “T” followed by any single character, followed by the letter “m,” and would retrieve rows that include the words Tim, Tom, or T3m.
ACCESS USER NOTE Access uses different wildcard symbols. The symbol for any collection of characters is the asterisk (*), as shown in Figure 4-18. The symbol for an individual character is the question mark (?).
SELECT CUSTOMER_NUM, CUSTOMER_NAME, STREET, CITY, STATE, ZIP FROM CUSTOMER WHERE STREET LIKE '*Central*';
FIGURE 4-18
Access SELECT command with wildcards
NOTE In a large database, you should use wildcards only when absolutely necessary. Searches involving wildcards can be extremely slow to process.
Using the IN Operator An IN clause, which consists of the IN operator followed by a collection of values, provides a concise way of phrasing certain conditions, as Example 13 illustrates. You will see another use for the IN clause in more complex examples later in this chapter.
EXAMPLE 13 List the number, name, and credit limit for each customer with a credit limit of $5,000, $10,000, or $15,000.
In this query, you can use an IN clause to determine whether a credit limit is $5,000, $10,000, or $15,000. You could obtain the same answer by using the condition WHERE Single-Table Queries
CREDIT_LIMIT = 5000 OR CREDIT_LIMIT = 10000 OR CREDIT_LIMIT = 15000. The approach shown in Figure 4-19 is simpler because the IN clause contains a collection of values: 5000, 10000, and 15000. The condition is true for those rows in which the value in the CREDIT_LIMIT column is in this collection. 112 List of values in the IN clause
FIGURE 4-19
SELECT command with an IN clause
SORTING Recall that the order of rows in a table is immaterial to the DBMS. From a practical standpoint, this means that when you query a relational database, there is no defined order in which to display the results. Rows might be displayed in the order in which the data was originally entered, but even this is not certain. If the order in which the data is displayed is important, you can specifically request that the results appear in a desired order. In SQL, you specify the results order by using the ORDER BY clause.
Using the ORDER BY Clause You use the ORDER BY clause to list data in a specific order, as shown in Example 14.
EXAMPLE 14 List the number, name, and balance of each customer. Order (sort) the output in ascending (increasing) order by balance.
The column on which to sort data is called a sort key or simply a key. In Example 14, you need to order the output by balance, so the sort key is the BALANCE column. To sort the output, use an ORDER BY clause followed by the sort key. If you do not specify a sort order, the default is ascending. The query appears in Figure 4-20.
Chapter 4
Sort key
113
Rows are sorted in ascending order by balance
FIGURE 4-20
SELECT command to sort rows
Additional Sorting Options Sometimes you might need to sort data using more than one key, as shown in Example 15.
EXAMPLE 15 List the number, name, and credit limit of each customer. Order the customers by name within descending credit limit. (In other words, first sort the customers by credit limit in descending order. Within each group of customers that have a common credit limit, sort the customers by name in ascending order.)
Example 15 involves two new ideas: sorting on multiple keys—CREDIT_LIMIT and CUSTOMER_NAME—and sorting one of the keys in descending order. When you need to sort data on two columns, the more important column (in this case, CREDIT_LIMIT) is called the major sort key (or the primary sort key) and the less important column (in this case, CUSTOMER_NAME) is called the minor sort key (or the secondary sort key). To sort on multiple keys, you list the keys in order of importance in the ORDER BY clause. To sort in descending order, you follow the name of the sort key with the DESC operator, as shown in Figure 4-21.
Single-Table Queries
Major sort key Minor sort key Descending order
114
Rows sorted in descending order by credit limit
Rows with the same credit limit are sorted by name
FIGURE 4-21
SELECT command to sort data using multiple sort keys
USING FUNCTIONS SQL uses special functions, called aggregate functions, to calculate sums, averages, counts, maximum values, and minimum values. These functions apply to groups of rows. They could apply to all the rows in a table (for example, calculating the average balance of all customers). They also could apply to those rows satisfying some particular condition (for example, the average balance of all customers of sales rep 20). The descriptions of the aggregate functions appear in Figure 4-22.
Function
Description
AVG
Calculates the average value in a column
COUNT
Determines the number of rows in a table
MAX
Determines the maximum value in a column
MIN
Determines the minimum value in a column
SUM
Calculates a total of the values in a column
FIGURE 4-22
SQL aggregate functions
Using the COUNT Function The COUNT function, as illustrated in Example 16, counts the number of rows in a table.
Chapter 4
EXAMPLE 16 How many parts are in item class HW?
For this query, you need to determine the total number of rows in the PART table with the value HW in the CLASS column. You could count the part numbers in the query results, or the number of part descriptions, or the number of entries in any other column. It doesn’t matter which column you choose because all columns should provide the same answer. Rather than arbitrarily selecting one column, most SQL implementations let you use the asterisk (*) to represent any column, as shown in Figure 4-23.
Condition
115
COUNT function in SELECT clause
Number of rows in the PART table on which the class is HW
FIGURE 4-23
SELECT command to count rows
You also can count the number of rows in a query by selecting a specific column instead of using the asterisk, as follows: SELECT COUNT(PART_NUM) FROM PART WHERE CLASS = 'HW';
Using the SUM Function If you need to calculate the total of all customers’ balances, you can use the SUM function, as illustrated in Example 17.
EXAMPLE 17 Find the total number of Premiere Products customers and the total of their balances.
When you use the SUM function, you must specify the column to total, and the column’s data type must be numeric. (How could you calculate a sum of names or addresses?) Figure 4-24 shows the query.
Single-Table Queries
116
FIGURE 4-24
SELECT command to count rows and calculate a total
Using the AVG, MAX, and MIN Functions Using the AVG, MAX, and MIN functions is similar to using SUM, except that different statistics are calculated. AVG calculates the average value in a numeric range, MAX calculates the maximum value in a numeric range, and MIN calculates the minimum value in a numeric range.
EXAMPLE 18 Find the sum of all balances, the average balance, the maximum balance, and the minimum balance of all Premiere Products customers.
Figure 4-25 shows the query and the results.
FIGURE 4-25
Chapter 4
SELECT command with several functions
NOTE When you use the SUM, AVG, MAX, or MIN functions, SQL ignores any null value(s) in the column and eliminates them from the computations. Null values in numeric columns can produce strange results when statistics are computed. Suppose the BALANCE column accepts null values, there are currently four customers in the CUSTOMER table, and their respective balances are $100, $200, $300, and null (unknown). When you calculate the average balance, SQL ignores the null value and obtains a result of $200 (($100 + $200 + $300) / 3). Similarly, when you calculate the total of the balances, SQL ignores the null value and calculates a total of $600. When you count the number of customers in the table, however, SQL includes the row containing the null value, and the result is 4. Thus the total of the balances ($600) divided by the number of customers (4) results in an average balance of $150!
117
NOTE You can use an AS clause with a function. For example, the following command computes a sum of the BALANCE column and displays the column heading as TOTAL_BALANCE in the query results: SELECT SUM(BALANCE) AS TOTAL_BALANCE FROM CUSTOMER;
Using the DISTINCT Operator In some situations, the DISTINCT operator is useful when used in conjunction with the COUNT function because it eliminates duplicate values in the query results. Examples 19 and 20 illustrate the most common uses of the DISTINCT operator.
EXAMPLE 19 Find the number of each customer that currently has an open order (that is, an order currently in the ORDERS table).
The command seems fairly simple. When a customer currently has an open order, there must be at least one row in the ORDERS table on which that customer’s number appears. You could use the query shown in Figure 4-26 to find the customer numbers with open orders.
Single-Table Queries
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Customer 608 has two orders on file
FIGURE 4-26
Numbers of customers with open orders
Notice that customer numbers 148 and 608 each appear more than once in the results; this means that both customers currently have more than one open order in the ORDERS table. Suppose you want to list each customer only once, as illustrated in Example 20.
EXAMPLE 20 Find the number of each customer that currently has an open order. List each customer only once.
To ensure uniqueness, you can use the DISTINCT operator, as shown in Figure 4-27.
FIGURE 4-27
Chapter 4
Numbers of customers with open orders and with duplicates removed
You might wonder about the relationship between COUNT and DISTINCT, because both involve counting rows. Example 21 identifies the differences.
EXAMPLE 21 119
Count the number of customers that currently have open orders.
The query shown in Figure 4-28 counts the number of customers using the CUSTOMER_NUM column.
FIGURE 4-28
Count that includes duplicate customer numbers
Q&A Question: What is wrong with the query results shown in Figure 4-28? Answer: The answer, 7, is the result of counting the customers that have open orders multiple times—once for each separate order currently on file. The result counts each customer number and does not eliminate duplicate customer numbers to provide an accurate count of the number of customers.
Some SQL implementations, including Oracle and SQL Server (but not Access), allow you to use the DISTINCT operator to calculate the correct count, as shown in Figure 4-29.
FIGURE 4-29
Count that excludes duplicate customer numbers (using DISTINCT within COUNT) Single-Table Queries
NESTING QUERIES Sometimes obtaining the results you need requires two or more steps, as shown in the next two examples. 120
EXAMPLE 22 List the number of each part in class AP.
The command to obtain the answer is shown in Figure 4-30.
FIGURE 4-30
Selecting all parts in class AP
EXAMPLE 23 List the order numbers that contain an order line for a part in class AP.
Example 23 asks you to find the order numbers in the ORDER_LINE table that correspond to the part numbers in the results of the query used in Example 22. After viewing those results (CD52, DR93, DW11, KL62, and KT03), you can use the command shown in Figure 4-31.
Chapter 4
Results from previous query
FIGURE 4-31
121
Query using the results from Figure 4-30
Subqueries It is possible to place one query inside another. The inner query is called a subquery. The subquery is evaluated first. After the subquery has been evaluated, the outer query can use the results of the subquery to find its results, as shown in Example 24.
EXAMPLE 24 Find the answer to Examples 22 and 23 in one step.
You can find the same result as in the previous two examples in a single step by using a subquery. In Figure 4-32, the command shown in parentheses is the subquery. This subquery is evaluated first, producing a temporary table. The temporary table is used only to evaluate the query—it is not available to the user or displayed—and it is deleted after the evaluation of the query is complete. In this example, the temporary table has only a single column (PART_NUM) and five rows (CD52, DR93, DW11, KL62, and KT03). The outer query is evaluated next. In this case, the outer query retrieves the order number on every row in the ORDER_LINE table for which the part number is in the results of the subquery. Because that table contains only the part numbers in class AP, the results display the desired list of order numbers.
Single-Table Queries
IN operator Subquery to find part numbers for parts in class AP
122
FIGURE 4-32
Using the IN operator and a subquery
Figure 4-32 shows duplicate order numbers in the results. To eliminate this duplication, you can use the DISTINCT operator as follows: SELECT DISTINCT(ORDER_NUM) FROM ORDER_LINE WHERE PART_NUM IN (SELECT PART_NUM FROM PART WHERE CLASS = 'AP');
The results of this query will display each order number only once.
EXAMPLE 25 List the number, name, and balance for each customer whose balance exceeds the average balance of all customers.
In this case, you use a subquery to obtain the average balance. Because this subquery produces a single number, you can compare each customer’s balance with this number, as shown in Figure 4-33.
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FIGURE 4-33
Query using an operator and a subquery
NOTE You cannot use the condition BALANCE > AVG(BALANCE) in the WHERE clause; you must use a subquery to obtain the average balance. Then you can use the results of the subquery in a condition, as illustrated in Figure 4-33.
GROUPING Grouping creates groups of rows that share some common characteristic. If you group customers by credit limit, for example, the first group contains customers with $5,000 credit limits, the second group contains customers with $7,500 credit limits, and so on. If, on the other hand, you group customers by sales rep number, the first group contains those customers represented by sales rep number 20, the second group contains those customers represented by sales rep number 35, and the third group contains those customers represented by sales rep number 65. When you group rows, any calculations indicated in the SELECT command are performed for the entire group. For example, if you group customers by rep number and the query requests the average balance, the results include the average balance for the group of customers represented by rep number 20, the average balance for the group represented by rep number 35, and the average balance for the group represented by rep number 65. The following examples illustrate this process.
Using the GROUP BY Clause The GROUP BY clause lets you group data on a particular column, such as REP_NUM, and then calculate statistics, when desired, as shown in Example 26.
Single-Table Queries
EXAMPLE 26 For each sales rep, list the rep number and the average balance of the rep’s customers.
124
Because you need to group customers by rep number and then calculate the average balance for all customers in each group, you must use the GROUP BY clause. In this case, GROUP BY REP_NUM puts customers with the same rep number into separate groups. Any statistics indicated in the SELECT command are calculated for each group. It is important to note that the GROUP BY clause does not sort the data in a particular order; you must use the ORDER BY clause to sort data. Assuming that the results should be ordered by rep number, you can use the command shown in Figure 4-34.
One row for each group
FIGURE 4-34
Grouping records on a column
When rows are grouped, one line of output is produced for each group. The only things that can be displayed are statistics calculated for the group or columns whose values are the same for all rows in a group.
Q&A Question: Is it appropriate to display the rep number in the query for Example 26? Answer: Yes, because the rep number in one row in a group must be the same as the rep number in any other row in the group.
Chapter 4
Q&A Question: Would it be appropriate to display a customer number in the query for Example 26? Answer: No, because the customer number varies on the rows in a group. (The same rep is associated with many customers.) The DBMS would not be able to determine which customer number to display for the group, and would display an error message if you attempt to display a customer number.
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Using a HAVING Clause The HAVING clause is used to restrict the groups that are included, as shown in Example 27.
EXAMPLE 27 Repeat the previous example, but list only those reps who represent fewer than four customers.
The only difference between Examples 26 and 27 is the restriction to display only those reps who represent fewer than four customers. This restriction does not apply to individual rows but rather to groups. Because the WHERE clause applies only to rows, you cannot use it to accomplish the kind of selection that is required. Fortunately, the HAVING clause does for groups what the WHERE clause does for rows. The HAVING clause limits the groups that are included in the results. In Figure 4-35, the row created for a group is displayed only when the count of the number of rows in the group is less than four; in addition, all groups are ordered by rep number.
Only groups with fewer than four rows are included
FIGURE 4-35
Restricting the groups to include in the results
HAVING vs. WHERE Just as you can use the WHERE clause to limit the rows that are included in a query’s result, you can use the HAVING clause to limit the groups that are included. The following examples illustrate the difference between these two clauses. Single-Table Queries
EXAMPLE 28 List each credit limit and the number of customers having each credit limit.
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To count the number of customers that have a given credit limit, you must group the data by credit limit, as shown in Figure 4-36.
FIGURE 4-36
Counting the number of rows in each group
EXAMPLE 29 Repeat Example 28, but list only those credit limits held by more than one customer.
Because this condition involves a group total, the query includes a HAVING clause, as shown in Figure 4-37.
FIGURE 4-37 Chapter 4
Displaying groups that contain more than one row
EXAMPLE 30 List each credit limit and the number of customers of sales rep 20 that have this limit.
The condition involves only rows, so using the WHERE clause is appropriate, as shown in Figure 4-38.
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Only rows on which the rep number is 20 are included in the groups
FIGURE 4-38
Restricting the rows to be grouped
EXAMPLE 31 Repeat Example 30, but list only those credit limits held by more than one customer.
Because the conditions involve rows and groups, you must use both a WHERE clause and a HAVING clause, as shown in Figure 4-39.
FIGURE 4-39
Restricting the rows and the groups
Single-Table Queries
In Example 31, rows from the original table are evaluated only when the sales rep number is 20. These rows then are grouped by credit limit and the count is calculated. Only groups for which the calculated count is greater than one are displayed.
NULLS 128
Sometimes a condition involves a column that can accept null values, as illustrated in Example 32.
EXAMPLE 32 List the number and name of each customer with a null (unknown) street value.
You might expect the condition to be something like STREET = NULL. The correct format actually uses the IS NULL operator (STREET IS NULL), as shown in Figure 4-40. (To select a customer whose street is not null, use the IS NOT NULL operator (STREET IS NOT NULL).) In the current Premiere Products database, no customer has a null street value; therefore, no rows are retrieved in the query results.
FIGURE 4-40
Selecting rows containing null values in the STREET column
SUMMARY OF SQL CLAUSES, FUNCTIONS, AND OPERATORS In this chapter, you learned how to create queries that retrieve data from a single table by constructing appropriate SELECT commands. In the next chapter, you will learn how to create queries that retrieve data from multiple tables. The queries you created in this chapter used the clauses, functions, and operators shown in Figure 4-41.
Chapter 4
Clause, function, or operator
Description
AND operator
Specifies that all simple conditions must be true for the compound condition to be true
AVG function
Calculates the average value in a numeric range
BETWEEN operator
Specifies a range of values in a condition
COUNT function
Counts the number of rows in a table
DESC operator
Sorts the query results in descending order based on the column name
DISTINCT operator
Ensures uniqueness in the condition by eliminating redundant values
FROM clause
Indicates the table from which to retrieve the specified columns
GROUP BY clause
Groups rows based on the specified column
HAVING clause
Limits a condition to the groups that are included
IN clause
Uses the IN operator to find a value in a group of values specified in the condition
IS NOT NULL operator
Finds rows that do not contain a null value in the specified column
IS NULL operator
Finds rows that contain a null value in the specified column
LIKE operator
Indicates a pattern of characters to find in a condition
MAX function
Calculates the maximum value in a numeric range
MIN function
Calculates the minimum value in a numeric range
NOT operator
Reverses the truth or falsity of the original condition
OR operator
Specifies that the compound condition is true whenever any of the simple conditions is true
ORDER BY clause
Lists the query results in the specified order based on the column name
SELECT clause
Specifies the columns to retrieve in the query
SUM function
Totals the values in a numeric range
WHERE clause
Specifies any conditions for the query
FIGURE 4-41
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SQL query clauses, functions, and operators
Single-Table Queries
Chapter Summary ●
The basic form of the SQL SELECT command is SELECT-FROM-WHERE. Specify the columns to be listed after the word SELECT (or type an asterisk (*) to select all columns), and then specify the table name that contains these columns after the word FROM. Optionally, you can include one or more conditions after the word WHERE.
●
Simple conditions are written in the following form: column name, comparison operator, column name or value. Simple conditions can involve any of the comparison operators: =, >, >=, <, <=, < >, or !=.
●
You can form compound conditions by combining simple conditions using the AND, OR, and NOT operators.
●
Use the BETWEEN operator to indicate a range of values in a condition.
●
Use computed columns in SQL commands by using arithmetic operators and writing the computation in place of a column name. You can assign a name to the computed column by following the computation with the word AS and then the desired name.
●
To check for a value in a character column that is similar to a particular string of characters, use the LIKE operator. In Oracle and SQL Server, the percent (%) wildcard represents any collection of characters, and the underscore ( _ ) wildcard represents any single character. In Access, the asterisk (*) wildcard represents any collection of characters, and the question mark (?) wildcard represents any single character.
●
To determine whether a column contains a value in a set of values, use the IN operator.
●
Use an ORDER BY clause to sort data. List sort keys in order of importance. To sort in descending order, follow the sort key with the DESC operator.
●
SQL processes the aggregate functions COUNT, SUM, AVG, MAX, and MIN. These calculations apply to groups of rows.
●
To avoid duplicates in a query that uses an aggregate function, precede the column name with the DISTINCT operator.
●
When one SQL query is placed inside another, it is called a subquery. The inner query (the subquery) is evaluated first.
●
Use a GROUP BY clause to group data.
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Use a HAVING clause to restrict the output to certain groups.
●
Use the IS NULL operator in a WHERE clause to find rows containing a null value in a particular column. Use the IS NOT NULL operator in a WHERE clause to find rows that do not contain a null value.
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Key Terms
Chapter 4
aggregate function
compound condition
AND
computed column
AND condition
COUNT
AVG
DESC
BETWEEN
DISTINCT
FROM clause
NOT condition
GROUP BY clause
OR
grouping
OR condition
HAVING clause
ORDER BY clause
IN clause
primary sort key
IS NOT NULL
query
IS NULL
secondary sort key
key
SELECT clause
LIKE
simple condition
major sort key
sort key
MAX
subquery
MIN
SUM
minor sort key
WHERE clause
131
NOT
Review Questions 1.
Describe the basic form of the SQL SELECT command.
2.
How do you form a simple condition?
3.
How do you form a compound condition?
4.
In SQL, what operator do you use to determine whether a value is between two other values without using an AND condition?
5.
How do you use a computed column in SQL? How do you name the computed column?
6.
In which clause would you use a wildcard in a condition?
7.
What wildcards are available in Oracle, and what do they represent?
8.
How do you determine whether a column contains one of a particular set of values without using an AND condition?
9.
How do you sort data?
10.
How do you sort data on more than one sort key? What is the more important key called? What is the less important key called?
11.
How do you sort data in descending order?
12.
What are the SQL aggregate functions?
13.
How do you avoid including duplicate values in a query’s results?
14.
What is a subquery?
15.
How do you group data in an SQL query?
16.
When grouping data in a query, how do you restrict the output to only those groups satisfying some condition?
17.
How do you find rows in which a particular column contains a null value?
Single-Table Queries
18.
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Use your favorite Web browser and Web search engine to find out how to enter a date in an SQL query in Oracle, Access, and SQL Server. Using the information you find, complete the following SQL command for each of the three DBMSs (Oracle, Access, and SQL Server) to list orders placed on October 20, 2010: SELECT * FROM ORDERS WHERE ORDER_DATE = Be sure to reference the URLs that contain the information.
Exercises Premiere Products Use SQL and the Premiere Products database (see Figure 1-2 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output.
Chapter 4
1.
List the part number, description, and price for all parts.
2.
List all rows and columns for the complete ORDERS table.
3.
List the names of customers with credit limits of $10,000 or more.
4.
List the order number for each order placed by customer number 608 on 10/23/2010. (Hint: If you need help, use the discussion of the DATE data type in Figure 3-11 in Chapter 3.)
5.
List the number and name of each customer represented by sales rep 35 or sales rep 65.
6.
List the part number and part description of each part that is not in item class AP.
7.
List the part number, description, and number of units on hand for each part that has between 10 and 25 units on hand, including both 10 and 25. Do this two ways.
8.
List the part number, part description, and on-hand value (units on hand * unit price) of each part in item class SG. (On-hand value is really units on hand * cost, but there is no COST column in the PART table.) Assign the name ON_HAND_VALUE to the computed column.
9.
List the part number, part description, and on-hand value for each part whose on-hand value is at least $7,500. Assign the name ON_HAND_VALUE to the computed column.
10.
Use the IN operator to list the part number and part description of each part in item class AP or SG.
11.
Find the number and name of each customer whose name begins with the letter “B.”
12.
List all details about all parts. Order the output by part description.
13.
List all details about all parts. Order the output by part number within warehouse. (That is, order the output by warehouse and then by part number.)
14.
How many customers have balances that are more than their credit limits?
15.
Find the total of the balances for all customers represented by sales rep 65 with balances that are less than their credit limits.
16.
List the part number, part description, and on-hand value of each part whose number of units on hand is more than the average number of units on hand for all parts. (Hint: Use a subquery.)
17.
What is the price of the least expensive part in the database?
18.
What is the part number, description, and price of the least expensive part in the database? (Hint: Use a subquery.)
19.
List the sum of the balances of all customers for each sales rep. Order and group the results by sales rep number.
20.
List the sum of the balances of all customers for each sales rep, but restrict the output to those sales reps for which the sum is more than $10,000.
21.
List the part number of any part with an unknown description.
133
Henry Books Use SQL and the Henry Books database (Figures 1-4 through 1-7 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
List the book code and book title of each book.
2.
List the complete BRANCH table.
3.
List the name of each publisher located in Boston.
4.
List the name of each publisher not located in New York.
5.
List the name of each branch that has at least nine employees.
6.
List the book code and book title of each book that has the type HOR.
7.
List the book code and book title of each book that has the type HOR and is in paperback.
8.
List the book code and book title of each book that has the type HOR or is published by the publisher with the publisher code SC.
9.
List the book code, book title, and price of each book with a price between $15 and $25.
10.
List the book code and book title of each book that has the type MYS and a price of less than $20.
11.
Customers who are part of a special program get a 10 percent discount off regular book prices. List the book code, book title, and discounted price of each book. Use DISCOUNTED_PRICE as the name for the computed column, which should calculate 90 percent of the current price (100 percent less a 10 percent discount).
12.
Find the name of each publisher containing the word “and.” (Hint: Be sure that your query selects only those publishers that contain the word “and” and not those that contain the letters “and” in the middle of a word. For example, your query should select the publisher named “Farrar Straus and Giroux,” but should not select the publisher named “Random House.”)
13.
List the book code and book title of each book that has the type SFI, MYS, or HOR. Use the IN operator in your command.
14.
Repeat Exercise 13, but also list the books in alphabetical order by title.
15.
Repeat Exercise 13, but also include the price, and list the books in descending order by price. Within a group of books having the same price, further order the books by title.
16.
Display the list of book types in the database. List each book type only once.
17.
How many books have the type SFI?
Single-Table Queries
134
18.
For each type of book, list the type and the average price.
19.
Repeat Exercise 18, but consider only paperback books.
20.
Repeat Exercise 18, but consider only paperback books for those types for which the average price is more than $10.
21.
What are the title(s) and price(s) of the least expensive book(s) in the database?
22.
What is the most expensive book in the database?
23.
How many employees does Henry Books have?
Alexamara Marina Group Use SQL and the Alexamara Marina Group database (Figures 1-8 through 1-12 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output.
Chapter 4
1.
List the owner number, last name, and first name of every boat owner.
2.
List the complete MARINA table (all rows and all columns).
3.
List the last name and first name of every owner who lives in Rivard.
4.
List the last name and first name of every owner who does not live in Rivard.
5.
List the marina number and slip number for every slip whose length is equal to or less than 30 feet.
6.
List the marina number and slip number for every boat with the type Ray 4025.
7.
List the slip number for every boat with the type Ray 4025 that is located in marina 1.
8.
List the boat name for each boat located in a slip whose length is between 25 and 30 feet.
9.
List the slip number for every slip in marina 1 whose rental fee is less than $3,000.
10.
Labor is billed at the rate of $60 per hour. List the slip ID, category number, estimated hours, and estimated labor cost for every service request. To obtain the estimated labor cost, multiply the estimated hours by 60. Use the column name ESTIMATED_COST for the estimated labor cost.
11.
List the marina number and slip number for all slips containing a boat with the type Sprite 4000, Sprite 3000, or Ray 4025.
12.
List the marina number, slip number, and boat name for all boats. Sort the results by boat name within marina number.
13.
How many Dolphin 28 boats are stored at both marinas?
14.
Calculate the total rental fees Alexamara receives each year based on the length of the slip.
CHAPTER
5
MULTIPLE-TABLE QUERIES LEARNING
OBJECTIVES
Objectives ●
Use joins to retrieve data from more than one table
●
Use the IN and EXISTS operators to query multiple tables
●
Use a subquery within a subquery
●
Use an alias
●
Join a table to itself
●
Perform set operations (union, intersection, and difference)
●
Use the ALL and ANY operators in a query
●
Perform special operations (inner join, outer join, and product)
INTRODUCTION In this chapter, you will learn how to use SQL to retrieve data from two or more tables using one SQL command. You will join tables together and examine how to obtain similar results using the SQL IN and EXISTS operators.Then you will use aliases to simplify queries and join a table to itself.You also will implement the set operations of union, intersection, and difference using SQL commands. You will examine two related SQL operators: ALL and ANY. Finally, you will perform inner joins, outer joins, and products.
Q U E R Y I N G M U L T I P L E TA B L E S In Chapter 4, you learned how to retrieve data from a single table. Many queries require you to retrieve data from two or more tables. To retrieve data from multiple tables, you first must join the tables, and then formulate a query using the same commands that you use for single-table queries.
NOTE In the following queries, your results might contain the same rows, but they might be listed in a different order. If order is important, you can include an ORDER BY clause in the query to ensure that the results are listed in the desired order.
136
Joining Two Tables To retrieve data from more than one table, you must join the tables together by finding rows in the two tables that have identical values in matching columns. You can join tables by using a condition in the WHERE clause, as you will see in Example 1.
EXAMPLE 1 List the number and name of each customer, together with the number, last name, and first name of the sales rep who represents the customer.
Because the customer numbers and names are in the CUSTOMER table and the sales rep numbers and names are in the REP table, you need to include both tables in the SQL command so you can retrieve data from both tables. To join (relate) the tables, you construct the SQL command as follows: 1. 2. 3.
In the SELECT clause, list all columns you want to display. In the FROM clause, list all tables involved in the query. In the WHERE clause, list the condition that restricts the data to be retrieved to only those rows from the two tables that match; that is, restrict it to the rows that have common values in matching columns.
As you learned in Chapter 2, it is often necessary to qualify a column name to specify the particular column you are referencing. Qualifying column names is especially important when joining tables because you must join tables on matching columns that frequently have identical column names. To qualify a column name, precede the name of the column with the name of the table, followed by a period. The matching columns in this example are both named REP_NUM—there is a column in the REP table named REP_NUM and a column in the CUSTOMER table that also is named REP_NUM. The REP_NUM column in the REP table is written as REP.REP_NUM and the REP_NUM column in the CUSTOMER table is written as CUSTOMER.REP_NUM. The query and its results appear in Figure 5-1.
Chapter 5
Condition to relate the tables
137
FIGURE 5-1
Joining two tables with a single SQL command
When there is potential ambiguity in listing column names, you must qualify the columns involved in the query. It is permissible to qualify other columns as well, even when there is no possible confusion. Some people prefer to qualify all column names; in this text, however, you will qualify column names only when necessary.
Q&A Question: In the first row of output in Figure 5-1, the customer number is 148, and the customer name is Al’s Appliance and Sport. These values represent the first row of the CUSTOMER table. Why is the sales rep number 20, the last name of the sales rep Kaiser, and the first name Valerie? Answer: In the CUSTOMER table, the sales rep number for customer number 148 is 20. (This indicates that customer number 148 is related to sales rep number 20.) In the REP table, the last name of sales rep number 20 is Kaiser and the first name is Valerie.
EXAMPLE 2 List the number and name of each customer whose credit limit is $7,500, together with the number, last name, and first name of the sales rep who represents the customer.
Multiple-Table Queries
In Example 1, you used a condition in the WHERE clause only to relate a customer with a sales rep to join the tables. Although relating a customer with a sales rep is essential in this example as well, you also need to restrict the output to only those customers whose credit limits are $7,500. You can restrict the rows by using a compound condition, as shown in Figure 5-2. 138 Condition to relate the tables Condition to restrict the rows
FIGURE 5-2
Restricting the rows in a join
EXAMPLE 3 For every part on order, list the order number, part number, part description, number of units ordered, quoted price, and unit price.
A part is considered “on order” when there is a row in the ORDER_LINE table in which the part appears. You can find the order number, number of units ordered, and quoted price in the ORDER_LINE table. To find the part description and the unit price, however, you need to look in the PART table. Then you need to find rows in the ORDER_LINE table and rows in the PART table that match (rows containing the same part number). The query and its results appear in Figure 5-3.
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139
FIGURE 5-3
Joining the ORDER_LINE and PART tables
Q&A Question: Can you use PART.PART_NUM instead of ORDER_LINE.PART_NUM in the SELECT clause? Answer: Yes. The values for these two columns match because they must satisfy the condition ORDER_LINE.PART_NUM = PART.PART_NUM.
COMPARING JOINS, IN, AND EXISTS You join tables in SQL by including a condition in the WHERE clause to ensure that matching columns contain equal values (for example, ORDER_LINE.PART_NUM = PART.PART_NUM). You can obtain similar results by using either the IN operator (described in Chapter 4) or the EXISTS operator with a subquery. The choice is a matter of personal preference because either approach obtains the same results. The following examples illustrate the use of each operator.
EXAMPLE 4 Find the description of each part included in order number 21610.
Because this query also involves retrieving data from the ORDER_LINE and PART tables (as illustrated in Example 3), you could approach it in a similar fashion. There are two basic differences, however, between Examples 3 and 4. First, the query in Example 4 does not require as many columns; second, it involves only order number 21610. Having fewer columns to retrieve means that there will be fewer columns listed in the SELECT clause. You
Multiple-Table Queries
can restrict the query to a single order by adding the condition ORDER_NUM = '21610' to the WHERE clause. The query and its results appear in Figure 5-4.
140
FIGURE 5-4
Restricting the rows when joining the ORDER_LINE and PART tables
Notice that the ORDER_LINE table is listed in the FROM clause, even though you do not need to display any columns from the ORDER_LINE table. The WHERE clause contains columns from the ORDER_LINE table, so it is necessary to include the table in the FROM clause.
Using the IN Operator Another way to retrieve data from multiple tables in a query is to use the IN operator with a subquery. In Example 4, you first could use a subquery to find all part numbers in the ORDER_LINE table that appear in any row on which the order number is 21610. Then you could find the part description for any part whose part number is in this list. The query and its results appear in Figure 5-5.
Outer query selects part descriptions in order 21610
Subquery selects part numbers in order 21610
FIGURE 5-5
Chapter 5
Using the IN operator instead of a join to query two tables
In Figure 5-5, evaluating the subquery produces a temporary table consisting of those part numbers (DR93 and DW11) that are present in order number 21610. Executing the remaining portion of the query produces part descriptions for each part whose number is in this temporary table; in this case, Gas Range (DR93) and Washer (DW11).
Using the EXISTS Operator You also can use the EXISTS operator to retrieve data from more than one table, as shown in Example 5. The EXISTS operator checks for the existence of rows that satisfy some criterion.
141
EXAMPLE 5 Find the order number and order date for each order that contains part number DR93.
This query is similar to the one in Example 4, but this time the query involves the ORDERS table and not the PART table. In this case, you can write the query in either of the ways previously demonstrated. For example, you could use the IN operator with a subquery, as shown in Figure 5-6.
FIGURE 5-6
Using the IN operator to select order information
Using the EXISTS operator provides another approach to solving Example 5, as shown in Figure 5-7.
Multiple-Table Queries
142
FIGURE 5-7
Using the EXISTS operator to select order information
The subquery in Figure 5-7 is the first one you have seen that involves a table listed in the outer query. This type of subquery is called a correlated subquery. In this case, the ORDERS table, which is listed in the FROM clause of the outer query, is used in the subquery. For this reason, you need to qualify the ORDER_NUM column in the subquery (ORDERS.ORDER_NUM). You did not need to qualify the columns in the previous queries involving the IN operator. The query shown in Figure 5-7 works as follows. For each row in the ORDERS table, the subquery is executed using the value of ORDERS.ORDER_NUM that occurs in that row. The inner query produces a list of all rows in the ORDER_LINE table in which ORDER_LINE.ORDER_NUM matches this value and in which PART_NUM is equal to DR93. You can precede a subquery with the EXISTS operator to create a condition that is true if one or more rows are obtained when the subquery is executed; otherwise, the condition is false. To illustrate the process, consider order numbers 21610 and 21613 in the ORDERS table. Order number 21610 is included because a row exists in the ORDER_LINE table with this order number and part number DR93. When the subquery is executed, there will be at least one row in the results, which in turn makes the EXISTS condition true. Order number 21613, however, will not be included because no row exists in the ORDER_LINE table with this order number and part number DR93. There will be no rows contained in the results of the subquery, which in turn makes the EXISTS condition false.
Using a Subquery Within a Subquery You can use SQL to create a nested subquery (a subquery within a subquery), as illustrated in Example 6.
EXAMPLE 6 Find the order number and order date for each order that includes a part located in warehouse 3.
Chapter 5
One way to approach this problem is first to determine the list of part numbers in the PART table for each part located in warehouse 3. Then you obtain a list of order numbers in the ORDER_LINE table with a corresponding part number in the part number list. Finally, you retrieve those order numbers and order dates in the ORDERS table for which the order number is in the list of order numbers obtained during the second step. The query and its results appear in Figure 5-8. 143 Outer query is evaluated last Intermediate query is evaluated second Innermost query is evaluated first
FIGURE 5-8
Nested subqueries (a subquery within a subquery)
As you might expect, SQL evaluates the queries from the innermost query to the outermost query. The query in this example is evaluated in three steps: 1. 2.
3.
The innermost subquery is evaluated first, producing a temporary table of part numbers for those parts located in warehouse 3. The next (intermediate) subquery is evaluated, producing a second temporary table with a list of order numbers. Each order number in this collection has a row in the ORDER_LINE table for which the part number is in the temporary table produced in Step 1. The outer query is evaluated last, producing the desired list of order numbers and order dates. Only those orders whose numbers are in the temporary table produced in Step 2 are included in the results.
Another approach to solving Example 6 involves joining the ORDERS, ORDER_LINE, and PART tables. The query and its results appear in Figure 5-9.
Multiple-Table Queries
144
FIGURE 5-9
Joining three tables
In this query, the following conditions join the tables: ORDER_LINE.ORDER_NUM = ORDERS.ORDER_NUM ORDER_LINE.PART_NUM = PART.PART_NUM
The condition WAREHOUSE = '3' restricts the output to only those parts located in warehouse 3. The query results are correct regardless of which command you use. You can use whichever approach you prefer. You might wonder whether one approach is more efficient than the other. SQL performs many built-in optimizations that analyze queries to determine the best way to satisfy them. Given a good optimizer, it should not make much difference how you formulate the query—you can see that using nested subqueries (Figure 5-8) produces the query in 0.11 seconds and joining the tables (Figure 5-9) produces the results in 0.02 seconds. If you are using a DBMS without an optimizer, however, the way you write a query can make a difference in the speed at which the DBMS executes the query. When you are working with a very large database and efficiency is a prime concern, consult the DBMS’s manual or try some timings yourself. Try running the same query both ways to see whether you notice a difference in the speed of execution. In small databases, there should not be a significant time difference between the two approaches.
A Comprehensive Example The query used in Example 7 involves several of the features already presented. The query illustrates all the major clauses that you can use in a SELECT command. It also illustrates the order in which these clauses must appear.
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EXAMPLE 7 List the customer number, order number, order date, and order total for each order with a total that exceeds $1,000. Assign the column name ORDER_TOTAL to the column that displays order totals. 145
The query and its results appear in Figure 5-10.
Name for computed column
FIGURE 5-10
Comprehensive example
In this query, the ORDERS and ORDER_LINE tables are joined by listing both tables in the FROM clause and relating them in the WHERE clause. Selected data is sorted by order number using the ORDER BY clause. The GROUP BY clause indicates that the data is to be grouped by order number, customer number, and order date. For each group, the SELECT clause displays the customer number, order number, order date, and order total (SUM(NUM_ORDERED * QUOTED_PRICE)). In addition, the total was renamed ORDER_TOTAL. Not all groups will be displayed, however. The HAVING clause displays only those groups whose SUM(NUM_ORDERED * QUOTED_PRICE) is greater than $1,000. The order number, customer number, and order date are unique for each order. Thus, it would seem that merely grouping by order number would be sufficient. SQL requires that both the customer number and the order date be listed in the GROUP BY clause. Recall that a SELECT clause can include statistics calculated for only the groups or columns whose values are identical for each row in a group. By stating that the data is to be grouped by order number, customer number, and order date, you tell SQL that the values in these columns must be the same for each row in a group.
Multiple-Table Queries
Using an Alias
146
When tables are listed in the FROM clause, you can give each table an alias, or an alternate name, that you can use in the rest of the statement. You create an alias by typing the name of the table, pressing the Spacebar, and then typing the name of the alias. No commas or periods are necessary to separate the two names. One reason for using an alias is simplicity. In Example 8, you assign the REP table the alias R and the CUSTOMER table the alias C. By doing this, you can type R instead of REP and C instead of CUSTOMER in the remainder of the query. The query in this example is simple, so you might not see the full benefit of this feature. When a query is complex and requires you to qualify the names, using aliases can simplify the process.
EXAMPLE 8 List the number, last name, and first name for each sales rep together with the number and name for each customer the sales rep represents.
The query and its results using aliases appear in Figure 5-11.
Alias
FIGURE 5-11
Using aliases in a query
NOTE Technically, it is unnecessary to qualify CUSTOMER_NUM because it is included only in the CUSTOMER table. It is qualified in Figure 5-11 for illustration purposes only.
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Joining a Table to Itself A second situation for using an alias is to join a table to itself, called a self-join, as illustrated in Example 9.
EXAMPLE 9 For each pair of customers located in the same city, display the customer number, customer name, and city.
147
If you had two separate tables for customers and the query requested customers in the first table having the same city as customers in the second table, you could use a normal join operation to find the answer. In this case, however, there is only one table (CUSTOMER) that stores all the customer information. You can treat the CUSTOMER table as if it were two tables in the query by creating an alias, as illustrated in Example 8. In this case, you use the following FROM clause: FROM CUSTOMER F, CUSTOMER S
SQL treats this clause as a query of two tables: one that has the alias F (first), and another that has the alias S (second). The fact that both tables are really the same CUSTOMER table is not a problem. The query and its results appear in Figure 5-12.
Condition to join F and S tables
FIGURE 5-12
Using aliases for a self-join
You are requesting a customer number and name from the F table, followed by a customer number and name from the S table, and then the city. (Because the city in the first table must match the city in the second table, you can select the city from either table.) The WHERE clause contains two conditions: the cities must match, and the customer number from the first table must be less than the customer number from the second table. In addition, the ORDER BY clause ensures that the data is sorted by the first customer
Multiple-Table Queries
number. For those rows with the same first customer number, the data is further sorted by the second customer number.
Q&A 148
Question: Why is the condition F.CUSTOMER_NUM < S.CUSTOMER_NUM important in the query? Answer: If you did not include this condition, you would get the query results shown in Figure 5-13.
FIGURE 5-13
Incorrect joining of a table to itself
The first row is included because it is true that customer number 148 (Al’s Appliance and Sport) in the F table has the same city as customer number 148 (Al’s Appliance and Sport) in the S table. The second row indicates that customer number 148 (Al’s Appliance and Sport) has the same city as customer number 524 (Kline’s). The eleventh row, however, repeats the same information because customer number 524 (Kline’s) has the same city as customer number 148 (Al’s Appliance and Sport). Of these three rows, the only row that should be included in the query results is the second row. The second row also is the only one of the three rows in which the first customer number (148) is less than the second customer number (524). This is why the query requires the condition F.CUSTOMER_NUM < S.CUSTOMER_NUM.
Using a Self-Join on a Primary Key Column Figure 5-14 shows some fields from an EMPLOYEE table whose primary key is EMPLOYEE_NUM. Another field in the table is MGR_EMPLOYEE_NUM, which represents Chapter 5
the number of the employee’s manager, who also is an employee. If you look at the row for employee 206 (Joan Dykstra), you will see that employee 198 (Mona Canzler) is Joan’s manager. By looking at the row for employee 198 (Mona Canzler), you see that her manager is employee 108 (Martin Holden). In the row for employee 108 (Martin Holden), the manager number is null, indicating that he has no manager. 149
Employee 108 has no manager Employee 198 manages employee 206
FIGURE 5-14
Employee and manager data
Suppose you need to list the employee number, employee last name, and employee first name along with the number, last name, and first name of each employee’s manager. Just as in the previous self-join, you would list the EMPLOYEE table twice in the FROM clause with aliases. The command shown in Figure 5-15 uses the letter E as an alias for the employee and the letter M as an alias for the manager. Thus E.EMPLOYEE_NUM is the employee’s number and M.EMPLOYEE_NUM is the number of the employee’s manager. In the SQL command, M.EMPLOYEE_NUM is renamed as MGR_NUM, M.LAST_NAME is renamed as MGR_LAST, and M.FIRST_NAME is renamed as MGR_FIRST. The condition in the WHERE clause ensures that E.MGR_EMPLOYEE_NUM (the number of the employee’s manager) matches M.EMPLOYEE_NUM (the employee number on the manager’s row in the table). Employee 108 is not included in the results because Martin Holden has no manager (see Figure 5-14).
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150
FIGURE 5-15
List of employees and their managers
Joining Several Tables It is possible to join several tables, as illustrated in Example 10. For each pair of tables you join, you must include a condition indicating how the columns are related.
EXAMPLE 10 For each part on order, list the part number, number ordered, order number, order date, customer number, and customer name, along with the last name of the sales rep who represents each customer.
A part is on order when it occurs on any row in the ORDER_LINE table. The part number, number ordered, and order number appear in the ORDER_LINE table. If these requirements represent the entire query, you would write the query as follows: SELECT PART_NUM, NUM_ORDERED, ORDER_NUM FROM ORDER_LINE;
This query is not sufficient, however. You also need the order date, which is in the ORDERS table; the customer number and name, which are in the CUSTOMER table; and the rep last name, which is in the REP table. Thus, you need to join four tables: ORDER_LINE, ORDERS, CUSTOMER, and REP. The procedure for joining more than two tables is essentially the same
Chapter 5
as the one for joining two tables. The difference is that the condition in the WHERE clause will be a compound condition. In this case, you would write the WHERE clause as follows: WHERE ORDERS.ORDER_NUM = ORDER_LINE.ORDER_NUM AND CUSTOMER.CUSTOMER_NUM = ORDERS.CUSTOMER_NUM AND REP.REP_NUM = CUSTOMER.REP_NUM
The first condition relates an order to an order line with a matching order number. The second condition relates the customer to the order with a matching customer number. The final condition relates the rep to a customer with a matching sales rep number. For the complete query, you list all the desired columns in the SELECT clause and qualify any columns that appear in more than one table. In the FROM clause, you list the tables that are involved in the query. The query and its results appear in Figure 5-16.
Conditions to relate the tables
FIGURE 5-16
151
Tables to include in query
Joining four tables in a query
Q&A Question: Why is the PART_NUM column, which appears in the PART and ORDER_LINE tables, not qualified in the SELECT clause? Answer: Among the tables listed in the query, only one table contains a column named PART_NUM, so it is not necessary to qualify the table. If the PART table also appeared in the FROM clause, you would need to qualify PART_NUM to avoid confusion between the PART_NUM columns in the PART and ORDER_LINE tables.
The query shown in Figure 5-16 is more complex than many of the previous ones you have examined. You might think that SQL is not such an easy language to use after all. If
Multiple-Table Queries
you take it one step at a time, however, the query in Example 10 really is not that difficult. To construct a detailed query in a step-by-step fashion, do the following: 1.
2. 152
3.
FIGURE 5-17
List in the SELECT clause all the columns that you want to display. If the name of a column appears in more than one table, precede the column name with the table name (that is, qualify the column name). List in the FROM clause all the tables involved in the query. Usually you include the tables that contain the columns listed in the SELECT clause. Occasionally, however, there might be a table that does not contain any columns used in the SELECT clause but that does contain columns used in the WHERE clause. In this case, you also must list the table in the FROM clause. For example, if you do not need to list a customer number or name, but you do need to list the rep name, you would not include any columns from the CUSTOMER table in the SELECT clause. The CUSTOMER table still is required, however, because you must include a column from it in the WHERE clause. Take one pair of related tables at a time and indicate in the WHERE clause the condition that relates the tables. Join these conditions with the AND operator. If there are any other conditions, include them in the WHERE clause and connect them to the other conditions with the AND operator. For example, if you want to view parts present on orders placed by only those customers with $10,000 credit limits, you would add one more condition to the WHERE clause, as shown in Figure 5-17.
Restricting the rows when joining four tables
SET OPERATIONS In SQL, you can use the set operations for taking the union, intersection, and difference of two tables. The union of two tables uses the UNION operator to create a temporary table containing every row that is in either the first table, the second table, or both tables. The intersection of two tables uses the INTERSECT operator to create a temporary table containing all rows that are in both tables. The difference of two tables uses the MINUS operator to create a temporary table containing the set of all rows that are in the first table but that are not in the second table. Chapter 5
For example, suppose that TEMP1 is a table containing the number and name of each customer represented by sales rep 65. Further suppose that TEMP2 is a table containing the number and name of those customers that currently have orders on file, as shown in Figure 5-18. TEMP1
TEMP2
CUSTOMER_NUM CUSTOMER_NAME
CUSTOMER_NUM CUSTOMER_NAME
356
Ferguson’s
148
Al’s Appliance and Sport
462
Bargains Galore
282
Brookings Direct
608
Johnson’s Department Store
356
Ferguson’s
408
The Everything Shop
608
Johnson’s Department Store
FIGURE 5-18
153
Customers of rep 65 and customers with open orders
The union of TEMP1 and TEMP2 (TEMP1 UNION TEMP2) consists of the number and name of those customers that are represented by sales rep 65 or that currently have orders on file, or both. The intersection of these two tables (TEMP1 INTERSECT TEMP2) contains those customers that are represented by sales rep 65 and that have orders on file. The difference of these two tables (TEMP1 MINUS TEMP2) contains those customers that are represented by sales rep 65 but that do not have orders on file. The results of these set operations are shown in Figure 5-19. TEMP1 UNION TEMP2
TEMP1 INTERSECT TEMP2
CUSTOMER_NUM CUSTOMER_NAME
CUSTOMER_NUM CUSTOMER_NAME
148
Al’s Appliance and Sport
356
Ferguson’s
282
Brookings Direct
608
Johnson’s Department Store
356
Ferguson’s
408
The Everything Shop
462
Bargains Galore
608
Johnson’s Department Store
TEMP1 MINUS TEMP2 CUSTOMER_NUM CUSTOMER_NAME
462
FIGURE 5-19
Bargains Galore
Union, intersection, and difference of the TEMP1 and TEMP2 tables
There is a restriction on set operations. It does not make sense, for example, to talk about the union of the CUSTOMER table and the ORDERS table because these tables do not contain the same columns. What might rows in this union look like? The two tables in the union must have the same structure for a union to be appropriate; the formal term is “union compatible.” Two tables are union compatible when they have the same number of columns and their corresponding columns have identical data types and lengths.
Multiple-Table Queries
Note that the definition of union compatible does not state that the columns of the two tables must be identical but rather that the columns must be of the same type. Thus, if one column is CHAR(20), the matching column also must be CHAR(20).
EXAMPLE 11 154
List the number and name of each customer that either is represented by sales rep 65 or that currently has orders on file, or both.
You can create a temporary table containing the number and name of each customer that is represented by sales rep 65 by selecting the customer numbers and names from the CUSTOMER table for which the sales rep number is 65. Then you can create another temporary table containing the number and name of each customer that currently has orders on file by joining the CUSTOMER and ORDERS tables. The two temporary tables created by this process have the same structure; that is, they both contain the CUSTOMER_NUM and CUSTOMER_NAME columns. Because the temporary tables are union compatible, it is possible to take the union of these two tables. The query and its results appear in Figure 5-20.
First query UNION operator
Second query
FIGURE 5-20
Using the UNION operator
If your SQL implementation truly supports the union operation, it will remove any duplicate rows automatically. For example, any customer that is represented by sales rep 65 and that currently has orders on file will appear only once in the results. Oracle, Access, and SQL Server support the union operation and correctly remove duplicates.
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EXAMPLE 12 List the number and name of each customer that is represented by sales rep 65 and that currently has orders on file.
The only difference between this query and the one in Example 11 is that the appropriate operator to use is INTERSECT, as shown in Figure 5-21.
FIGURE 5-21
155
Using the INTERSECT operator
Some SQL implementations do not support the INTERSECT operator, so you need to take a different approach. The command shown in Figure 5-22 produces the same results as the INTERSECT operator by using the IN operator and a subquery. The command selects the number and name of each customer that is represented by sales rep 65 and whose customer number also appears in the collection of customer numbers in the ORDERS table.
Rep number must be 65
Customer number must be in the results of the subquery Subquery to select numbers of customers with orders
FIGURE 5-22
Performing an intersection without using the INTERSECT operator
Multiple-Table Queries
NOTE Oracle and SQL Server support the INTERSECT operator but Microsoft Access does not.
EXAMPLE 13 156
List the number and name of each customer that is represented by sales rep 65 but that does not have orders currently on file.
The query uses the MINUS operator, as shown in Figure 5-23.
FIGURE 5-23
Using the MINUS operator
Just as with the INTERSECT operator, some SQL implementations do not support the MINUS operator. In such cases, you need to take a different approach, such as the one shown in Figure 5-24. This command produces the same results by selecting the number and name of each customer that is represented by sales rep 65 and whose customer number does not appear in the collection of customer numbers in the ORDERS table.
Customer number cannot be in the subquery results
FIGURE 5-24 Chapter 5
Performing a difference without using the MINUS operator
NOTE Oracle supports the MINUS operator, but SQL Server and Microsoft Access do not.
ALL AND ANY You can use the ALL and ANY operators with subqueries to produce a single column of numbers. When you precede the subquery by the ALL operator, the condition is true only if it satisfies all values produced by the subquery. When you precede the subquery by the ANY operator, the condition is true only if it satisfies any value (one or more) produced by the subquery. The following examples illustrate the use of these operators.
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EXAMPLE 14 Find the customer number, name, current balance, and rep number of each customer whose balance exceeds the maximum balance of all customers represented by sales rep 65.
You can find the maximum balance of the customers represented by sales rep 65 in a subquery and then find all customers whose balances are greater than this number. There is an alternative method that is simpler, however. You can use the ALL operator, as shown in Figure 5-25.
ALL operator
FIGURE 5-25
SELECT command that uses the ALL operator
To some users, the query shown in Figure 5-25 might seem more natural than finding the maximum balance in the subquery. For other users, the opposite might be true. You can use whichever approach you prefer.
Multiple-Table Queries
Q&A Question: How would you get the same result for Example 14 without using the ALL operator? Answer: You could select each customer whose balance is greater than the maximum balance of any customer of sales rep 65, as shown in Figure 5-26. 158
FIGURE 5-26
Alternative to using the ALL operator
EXAMPLE 15 Find the customer number, name, current balance, and rep number of each customer whose balance is greater than the balance of at least one customer of sales rep 65.
You can find the minimum balance of the customers represented by sales rep 65 in a subquery and then find all customers whose balance is greater than this number. To simplify the process, you can use the ANY operator, as shown in Figure 5-27.
Chapter 5
ANY operator
159
FIGURE 5-27
SELECT command with an ANY operator
Q&A Question: How would you get the same results without using the ANY operator? Answer: You could select each customer whose balance is greater than the minimum balance of any customer of sales rep 65, as shown in Figure 5-28.
Multiple-Table Queries
160
FIGURE 5-28
Alternative to using the ANY operator
SPECIAL OPERATIONS You can perform special operations within SQL, such as the self-join that you already used. Three other special operations are the inner join, the outer join, and the product.
Inner Join A join that compares the tables in a FROM clause and lists only those rows that satisfy the condition in the WHERE clause is called an inner join. The joins that you have performed so far in this text have been inner joins. Example 16 illustrates the inner join.
EXAMPLE 16 Display the customer number, customer name, order number, and order date for each order. Sort the results by customer number.
This example requires the same type of join that you have been using. The command is: SELECT CUSTOMER.CUSTOMER_NUM, CUSTOMER_NAME, ORDER_NUM, ORDER_DATE FROM CUSTOMER, ORDERS WHERE CUSTOMER.CUSTOMER_NUM = ORDERS.CUSTOMER_NUM ORDER BY CUSTOMER.CUSTOMER_NUM;
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The previous approach should work in any SQL implementation. An update to the SQL standard approved in 1992, called SQL-92, provides an alternative way of performing an inner join, as demonstrated in Figure 5-29.
INNER JOIN clause
161
ON clause
FIGURE 5-29
Query that uses an INNER JOIN clause
In the FROM clause, list the first table, and then include an INNER JOIN clause that includes the name of the second table. Instead of a WHERE clause, use an ON clause containing the same condition that you would have included in the WHERE clause.
Outer Join Sometimes you need to list all the rows from one of the tables in a join, regardless of whether they match any rows in a second table. For example, you can perform the join of the CUSTOMER and ORDERS tables in the query for Example 16, but display all customers—even the ones without orders. This type of join is called an outer join. There are actually three types of outer joins. In a left outer join, all rows from the table on the left (the table listed first in the query) are included regardless of whether they match rows from the table on the right (the table listed second in the query). Rows from the table on the right are included only when they match. In a right outer join, all rows from the table on the right are included regardless of whether they match rows from the table on the left. Rows from the table on the left are included only when they match. In a full outer join, all rows from both tables are included regardless of whether they match rows from the other table. (The full outer join is rarely used.) Example 17 illustrates the use of a left outer join.
Multiple-Table Queries
EXAMPLE 17 Display the customer number, customer name, order number, and order date for all orders. Include all customers in the results. For customers that do not have orders, omit the order number and order date. 162
To include all customers, you must perform an outer join. Assuming the CUSTOMER table is listed first, the join should be a left outer join. In SQL, you use the LEFT JOIN clause to perform a left outer join as shown in Figure 5-30. (You would use a RIGHT JOIN clause to perform a right outer join.)
LEFT JOIN clause
Customers without matching orders are also included
FIGURE 5-30
Query that uses a LEFT JOIN clause
All customers are included in the results. For customers without orders, the order number and date are blank. Technically, these blank values are null.
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NOTE In Oracle, there is another way to perform left and right outer joins. You write the join as you have been doing, with one exception. You include parentheses and a plus sign in the WHERE clause after the column in the table for which only matching rows are to be included. In this example, the plus sign would follow the CUSTOMER_NUM column in the ORDERS table because only orders that match customers are to be included. Because customers that do not have orders are to be included in the results, there would be no plus sign after the CUSTOMER_NUM column in the CUSTOMER table. The correct query is as follows:
163
SELECT CUSTOMER.CUSTOMER_NUM, CUSTOMER_NAME, ORDER_NUM, ORDER_DATE FROM CUSTOMER, ORDERS WHERE CUSTOMER.CUSTOMER_NUM = ORDERS.CUSTOMER_NUM(+) ORDER BY CUSTOMER.CUSTOMER_NUM; Running this query produces the same results shown in Figure 5-30.
Product The product (formally called the Cartesian product) of two tables is the combination of all rows in the first table and all rows in the second table.
NOTE The product operation is not common. You need to be aware of it, however, because it is easy to create a product inadvertently by omitting the WHERE clause when you are attempting to join tables.
EXAMPLE 18 Form the product of the CUSTOMER and ORDERS tables. Display the customer number and name from the CUSTOMER table, along with the order number and order date from the ORDERS table.
Forming a product is actually very easy. You simply omit the WHERE clause, as shown in Figure 5-31.
Multiple-Table Queries
No condition relates the tables in the FROM clause
164
FIGURE 5-31
Query that produces a product of two tables
Q&A Question: Figure 5-31 does not show all the rows in the result. How many rows are actually included? Answer: The CUSTOMER table has 10 rows and the ORDERS table has seven rows. Because each of the 10 customer rows is matched with each of the seven order rows, there are 70 (10 x 7) rows in the result.
Chapter 5
Chapter Summary ●
To join tables, indicate in the SELECT clause all columns to display, list in the FROM clause all tables to join, and then include in the WHERE clause any conditions requiring values in matching columns to be equal.
●
When referring to matching columns in different tables, you must qualify the column names to avoid confusion. You qualify column names using the following format: table name.column name.
●
Use the IN or EXISTS operators with an appropriate subquery as an alternate way of performing a join.
●
A subquery can contain another subquery. The innermost subquery is executed first.
●
The name of a table in a FROM clause can be followed by an alias, which is an alternate name for a table. You can use the alias in place of the table name throughout the SQL command. By using two different aliases for the same table in a single SQL command, you can join a table to itself.
●
The UNION operator creates a union of two tables (the collection of rows that are in either or both tables). The INTERSECT operator creates the intersection of two tables (the collection of rows that are in both tables). The MINUS operator creates the difference of two tables (the collection of rows that are in the first table but not in the second table). To perform any of these operations, the tables involved must be union compatible. Two tables are union compatible when they have the same number of columns and their corresponding columns have identical data types and lengths.
●
When the ALL operator precedes a subquery, the condition is true only if it is satisfied by all values produced by the subquery.
●
When the ANY operator precedes a subquery, the condition is true only if it is satisfied by any value (one or more) produced by the subquery.
●
In an inner join, only matching rows from both tables are included. You can use the INNER JOIN clause to perform an inner join.
●
In a left outer join, all rows from the table on the left (the table listed first in the query) are included regardless of whether they match rows from the table on the right (the table listed second in the query). Rows from the table on the right are included only when they match. You can use the LEFT JOIN clause to perform a left outer join. In a right outer join, all rows from the table on the right are included regardless of whether they match rows from the table on the left. Rows from the table on the left are included only when they match. You can use the RIGHT JOIN clause to perform a right outer join.
●
The product (Cartesian product) of two tables is the combination of all rows in the first table and all rows in the second table. To form a product of two tables, include both tables in the FROM clause and omit the WHERE clause.
165
Multiple-Table Queries
Key Terms
166
alias
join
ALL
left outer join
ANY
MINUS
Cartesian product
nested subquery
correlated subquery
outer join
difference
product
EXISTS
right outer join
full outer join
self-join
inner join
union
INTERSECT
UNION
intersection
union compatible
Review Questions
Chapter 5
1.
How do you join tables in SQL?
2.
When must you qualify names in SQL commands? How do you qualify a column name?
3.
List two operators that you can use with subqueries as an alternate way of performing joins.
4.
What is a nested subquery? In which order does SQL evaluate nested subqueries?
5.
What is an alias? How do you specify an alias in SQL? How do you use an alias?
6.
How do you join a table to itself in SQL?
7.
How do you take the union of two tables in SQL? How do you take the intersection of two tables in SQL? How do you take the difference of two tables in SQL? Are there any restrictions on the tables when performing any of these operations?
8.
What does it mean for two tables to be union compatible?
9.
How do you use the ALL operator with a subquery?
10.
How do you use the ANY operator with a subquery?
11.
Which rows are included in an inner join? What clause can you use to perform an inner join in SQL?
12.
Which rows are included in a left outer join? What clause can you use to perform a left outer join in SQL?
13.
Which rows are included in a right outer join? What clause can you use to perform a right outer join in SQL?
14.
What is the formal name for the product of two tables? How do you form a product in SQL?
15.
Use your favorite Web browser and Web search engine to find definitions for the terms equi-join, natural join, and cross join. Write a short report that identifies how these terms relate to the terms join, inner join, and Cartesian product. Be sure to reference your online sources properly.
16.
Use your favorite Web browser and Web search engine to find information on cost-based query optimizers. Write a short report that explains how cost-based query optimization works, and what type(s) of queries benefit the most from cost-based query optimization. Be sure to reference your online sources properly.
Exercises 167
Premiere Products Use SQL and the Premiere Products database (see Figure 1-2 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
For each order, list the order number and order date along with the number and name of the customer that placed the order.
2.
For each order placed on October 23, 2010, list the order number along with the number and name of the customer that placed the order.
3.
For each order, list the order number, order date, part number, number of units ordered, and quoted price for each order line that makes up the order.
4.
Use the IN operator to find the number and name of each customer that placed an order on October 23, 2010.
5.
Repeat Exercise 4, but this time use the EXISTS operator in your answer.
6.
Find the number and name of each customer that did not place an order on October 23, 2010.
7.
For each order, list the order number, order date, part number, part description, and item class for each part that makes up the order.
8.
Repeat Exercise 7, but this time order the rows by item class and then by order number.
9.
Use a subquery to find the rep number, last name, and first name of each sales rep who represents at least one customer with a credit limit of $10,000. List each sales rep only once in the results.
10.
Repeat Exercise 9, but this time do not use a subquery.
11.
Find the number and name of each customer that currently has an order on file for a Gas Range.
12.
List the part number, part description, and item class for each pair of parts that are in the same item class. (For example, one such pair would be part AT94 and part FD21, because the item class for both parts is HW.)
13.
List the order number and order date for each order placed by the customer named Johnson’s Department Store. (Hint: To enter an apostrophe (single quotation mark) within a string of characters, type two single quotation marks.)
14.
List the order number and order date for each order that contains an order line for an Iron.
15.
List the order number and order date for each order that either was placed by Johnson’s Department Store or that contains an order line for a Gas Range.
16.
List the order number and order date for each order that was placed by Johnson’s Department Store and that contains an order line for a Gas Range. Multiple-Table Queries
17.
List the order number and order date for each order that was placed by Johnson’s Department Store but that does not contain an order line for a Gas Range.
18.
List the part number, part description, unit price, and item class for each part that has a unit price greater than the unit price of every part in item class AP. Use either the ALL or ANY operator in your query. (Hint: Make sure you select the correct operator.)
19.
If you used ALL in Exercise 18, repeat the exercise using ANY. If you used ANY, repeat the exercise using ALL, and then run the new command. What question does this command answer?
20.
For each part, list the part number, description, units on hand, order number, and number of units ordered. All parts should be included in the results. For those parts that are currently not on order, the order number and number of units ordered should be left blank. Order the results by part number.
168
Henry Books Use SQL and the Henry Books database (see Figures 1-4 through 1-7 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output.
Chapter 5
1.
For each book, list the book code, book title, publisher code, and publisher name. Order the results by publisher name.
2.
For each book published by Scribner, list the book code, book title, and price.
3.
List the book title, book code, and price of each book published by Scribner that has a book price of at least $14.
4.
List the book code, book title, and units on hand for each book in branch number 3.
5.
List the book title for each book that has the type PSY and that is published by Berkley Publishing.
6.
Find the book title for each book written by author number 18. Use the IN operator in your query.
7.
Repeat Exercise 6, but this time use the EXISTS operator in your query.
8.
Find the book code and book title for each book located in branch number 2 and written by author 20.
9.
List the book codes for each pair of books that have the same price. (For example, one such pair would be book 0200 and book 7559, because the price of both books is $8.00.) The first book code listed should be the major sort key, and the second book code should be the minor sort key.
10.
Find the book title, author last name, and units on hand for each book in branch number 4.
11.
Repeat Exercise 10, but this time list only paperback books.
12.
Find the book code and book title for each book whose price is more than $10 or that was published in Boston.
13.
Find the book code and book title for each book whose price is more than $10 and that was published in Boston.
14.
Find the book code and book title for each book whose price is more than $10 but that was not published in Boston.
15.
Find the book code and book title for each book whose price is greater than the book price of every book that has the type MYS.
16.
Find the book code and book title for each book whose price is greater than the price of at least one book that has the type MYS.
17.
List the book code, book title, and units on hand for each book in branch number 2. Be sure each book is included, regardless of whether there are any copies of the book currently on hand in branch 2. Order the output by book code.
169
Alexamara Marina Group Use SQL and the Alexamara Marina Group database (see Figures 1-8 through 1-12 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
For every boat, list the marina number, slip number, boat name, owner number, owner’s first name, and owner’s last name.
2.
For every completed or open service request for routine engine maintenance, list the slip ID, description, and status.
3.
For every service request for routine engine maintenance, list the slip ID, marina number, slip number, estimated hours, spent hours, owner number, and owner’s last name.
4.
List the first and last names of all owners who have a boat in a 30-foot slip. Use the IN operator in your query.
5.
Repeat Exercise 4, but this time use the EXISTS operator in your query.
6.
List the names of any pair of boats that have the same type. For example, one pair would be Anderson II and Escape, because the boat type for both boats is Sprite 4000. The first name listed should be the major sort key and the second name should be the minor sort key.
7.
List the boat name, owner number, owner last name, and owner first name for each boat in marina 1.
8.
Repeat Exercise 7, but this time only list boats in 40-foot slips.
9.
List the marina number, slip number, and boat name for boats whose owners live in Glander Bay or whose type is Sprite 4000.
10.
List the marina number, slip number, and boat name for boats whose owners live in Glander Bay and whose type is Sprite 4000.
11.
List the marina number, slip number, and boat name for boats whose owners live in Glander Bay but whose type is not Sprite 4000.
12.
Find the service ID and slip ID for each service request whose estimated hours is greater than the number of estimated hours of at least one service request on which the category number is 3.
13.
Find the service ID and slip ID for each service request whose estimated hours is greater than the number of estimated hours on every service request on which the category number is 3.
14.
List the slip ID, boat name, owner number, service ID, number of estimated hours, and number of spent hours for each service request on which the category number is 2.
15.
Repeat Exercise 14, but this time be sure each slip is included regardless of whether the boat in the slip currently has any service requests for category 2. Multiple-Table Queries
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CHAPTER
6
U P D ATIN G D ATA LEARNING
OBJECTIVES
Objectives ●
Create a new table from an existing table
●
Change data using the UPDATE command
●
Add new data using the INSERT command
●
Delete data using the DELETE command
●
Use nulls in an UPDATE command
●
Change the structure of an existing table
●
Use the COMMIT and ROLLBACK commands to make permanent data updates or to reverse updates
●
Understand transactions and the role of COMMIT and ROLLBACK in supporting transactions
●
Drop a table
INTRODUCTION In this chapter, you will learn how to create a new table from an existing table and make changes to the data in a table.You will use the UPDATE command to change data in one or more rows in a table, and use the INSERT command to add new rows. You will use the DELETE command to delete rows. You will learn how to change the structure of a table in a variety of ways and use nulls in update operations.You will use the COMMIT command to make changes permanent and use the ROLLBACK command to undo changes, and understand how to use these commands in transactions. Finally, you will learn how to delete a table and its data.
C R E A T I N G A N EW TA B L E F RO M A N E X I S T I N G TA B L E You can create a new table using data in an existing table, as illustrated in the following examples.
EXAMPLE 1 172
Create a new table named LEVEL1_CUSTOMER that contains the following columns from the CUSTOMER table: CUSTOMER_NUM, CUSTOMER_NAME, BALANCE, CREDIT_LIMIT, and REP_NUM. The columns in the new LEVEL1_CUSTOMER table should have the same characteristics as the corresponding columns in the CUSTOMER table.
You describe the new table named LEVEL1_CUSTOMER by using the CREATE TABLE command shown in Figure 6-1.
Table to create Table structure
FIGURE 6-1
Creating the LEVEL1_CUSTOMER table
ACCESS USER NOTE If you are using Access to create the LEVEL1_CUSTOMER table, use the CURRENCY data type instead of the DECIMAL data type for the BALANCE and CREDIT_LIMIT fields. (Access does not support the DECIMAL data type.) You do not need to enter the field size and number of decimal places when using the CURRENCY data type.
EXAMPLE 2 Insert into the LEVEL1_CUSTOMER table the customer number, customer name, balance, credit limit, and rep number for customers with credit limits of $7,500.
Chapter 6
You can create a SELECT command to select the desired data from the CUSTOMER table, just as you did in Chapter 4. By placing this SELECT command in an INSERT command, you can add the query results to a table. The INSERT command appears in Figure 6-2; this command inserts four rows into the LEVEL1_CUSTOMER table.
173
SELECT command retrieves the data to insert
FIGURE 6-2
INSERT command to add data to the LEVEL1_CUSTOMER table
The SELECT command shown in Figure 6-3 displays the data in the LEVEL1_CUSTOMER table. Notice that the data comes from the new table you just created (LEVEL1_CUSTOMER), and not from the CUSTOMER table.
FIGURE 6-3
LEVEL1_CUSTOMER data
C H A N G I N G E X I S T I N G D A T A I N A TA B L E The data stored in tables is subject to constant change; prices, addresses, commission amounts, and other data in a database change on a regular basis. To keep data current, you must be able to make these changes to the data in your tables. You can use the UPDATE command to change rows for which a specific condition is true.
Updating Data
EXAMPLE 3 Change the name of customer 842 in the LEVEL1_CUSTOMER table to “All Season Sport.”
174
The format for the UPDATE command is the word UPDATE, followed by the name of the table to be updated. The next portion of the command consists of the word SET, followed by the name of the column to be updated, an equals sign, and the new value. When necessary, include a WHERE clause to indicate the row(s) on which the change is to occur. The UPDATE command shown in Figure 6-4 changes the name of customer 842 to All Season Sport.
FIGURE 6-4
UPDATE command to change the name of customer 842
The SELECT command shown in Figure 6-5 shows the data in the table after the change has been made. It is a good idea to use a SELECT command to display the data you changed to verify that the correct update was made.
FIGURE 6-5
Chapter 6
LEVEL1_CUSTOMER table after update
EXAMPLE 4 For each customer in the LEVEL1_CUSTOMER table that is represented by sales rep 20 and also has a balance that does not exceed the credit limit, increase the customer’s credit limit to $8,000.
The only difference between Examples 3 and 4 is that Example 4 uses a compound condition to identify the row(s) to be changed. The UPDATE command appears in Figure 6-6.
FIGURE 6-6
175
Using a compound condition in an update
The SELECT command shown in Figure 6-7 shows the table after the update.
FIGURE 6-7
Credit limit increased for customer number 148
You also can use the existing value in a column and a calculation to update a value. For example, when you need to increase the credit limit by 10 percent instead of changing it to a specific value, you can multiply the existing credit limit by 1.10. The following SET clause makes this change: SET CREDIT_LIMIT = CREDIT_LIMIT * 1.10
Updating Data
ADDING NEW ROWS TO AN EXISTING TA B L E In Chapter 3, you used the INSERT command to add the initial rows to the tables in the database. You also can use the INSERT command to add additional rows to tables.
EXAMPLE 5 176
Add customer number 895 to the LEVEL1_CUSTOMER table. The name is Peter and Margaret’s, the balance is 0, the credit limit is $8,000, and the rep number is 20.
The appropriate INSERT command is shown in Figure 6-8. Because the name “Peter and Margaret’s” contains an apostrophe, you type two single quotation marks to create the apostrophe.
Check mark removed from Autocommit check box Values for new row
FIGURE 6-8
Inserting a row
The SELECT command in Figure 6-9 shows that the row was successfully added.
FIGURE 6-9 Chapter 6
Customer 895 added to LEVEL1_CUSTOMER table
NOTE Your output might be sorted in a different order from what is shown in Figure 6-9. If you need to sort the rows in a specific order, use an ORDER BY clause with the desired sort key(s).
COMMIT AND ROLLBACK Figure 6-8 shows that the user cleared the check mark from the Autocommit check box before running the query. Autocommit is the default transaction mode and commits (makes permanent) each action query (INSERT, UPDATE, DELETE) as soon as the user executes the query. Although the Autocommit transaction mode is fine for most action queries, there are times when the user needs better control over when a transaction is committed. This is particularly important in multi-user database applications when more than one person can update the database and in applications when users are running script files that contain multiple updates. When you need more control over when transactions are committed, you should disable the Autocommit feature by clearing its check box before executing a query. If you do not use Autocommit, queries that include updates to table data are only temporary and you can reverse (cancel) them at any time during your current work session. Updates become permanent automatically when you exit from the DBMS. If you are not using Autocommit during your current work session, however, you can still commit (save) your changes immediately by executing the COMMIT command. If you decide that you do not want to save the changes you have made during your current work session, you can roll back (reverse) the changes by executing the ROLLBACK command. Any updates made since you ran the most recent COMMIT command will be reversed when you run the ROLLBACK command. If you have not run the COMMIT command, executing the ROLLBACK command will reverse all updates made during the current work session. You should note that the ROLLBACK command reverses only changes made to the data; it does not reverse changes made to a table’s structure. For example, if you change the length of a character column, you cannot use the ROLLBACK command to return the column length to its original state. If you determine that an update was made incorrectly, you can use the ROLLBACK command to return the data to its original state. If, on the other hand, you have verified that the update you made is correct, you can use the COMMIT command to make the update permanent. You do this by typing COMMIT; after running the update. However, you should note that the COMMIT command is permanent; after executing a COMMIT command, running the ROLLBACK command cannot reverse the update.
177
ACCESS USER NOTE Access does not support the COMMIT or ROLLBACK commands.
Updating Data
SQL SERVER USER NOTE In SQL Server, the commands used to commit and roll back data are COMMIT TRANSACTION and ROLLBACK TRANSACTION. By default, SQL Server is in Autocommit transaction mode. To turn off the Autocommit feature, execute the following command: SET XACT_ABORT ON To turn the Autocommit feature back on, execute the following command: SET XACT_ABORT OFF 178
TR A N S A C T I O N S A transaction is a logical unit of work. You can think of a transaction as a sequence of steps that accomplish a single task. When discussing transactions, it is essential that the entire sequence is completed successfully. For example, to enter an order, you must add the corresponding order to the ORDERS table, and then add each order line in the order to the ORDER_LINE table. These multiple steps accomplish the “single” task of entering an order. Suppose you have added the order and the first order line, but you are unable to enter the second order line for some reason; perhaps the part on the order line does not exist. This problem would leave the order in a partially entered state, which is unacceptable. To prevent this problem, you would execute a rollback, thus reversing the insertion of the order and the first order line. You can use the COMMIT and ROLLBACK commands to support transactions as follows: ●
●
●
Before beginning the updates for a transaction, commit any previous updates by executing the COMMIT command. Complete the updates for the transaction. If any update cannot be completed, execute the ROLLBACK command and discontinue the updates for the current transaction. If you can complete all updates successfully, execute the COMMIT command after completing the final update.
CHANGING AND DELETING EXISTING ROWS As you learned in Chapter 3, you use the DELETE command to remove rows from a table. In Example 6, you will change data and then use the DELETE command to delete a customer from the LEVEL1_CUSTOMER table. In Example 7, you will execute a rollback to reverse the updates made in Example 6. In this case, the rollback will return the row to its previous state and reinstate the deleted record.
EXAMPLE 6 In the LEVEL1_CUSTOMER table, change the name of customer 356 to “Smith Sport,” and then delete customer 895.
Chapter 6
To delete data from the database, use the DELETE command. The format for the DELETE command is the word DELETE followed by the name of the table containing the row(s) to be deleted. Next, use a WHERE clause with a condition to select the row(s) to delete. All rows satisfying the condition will be deleted. The first part of Example 6 requests a name change for customer 356; the command shown in Figure 6-10 makes this change.
179
FIGURE 6-10
Using an UPDATE command to change the name of customer 356
The second part of Example 6 requires deleting customer 895; this command is shown in Figure 6-11.
FIGURE 6-11
Using a DELETE command to delete customer 895
The command shown in Figure 6-12 displays the data in the table, verifying the change and the deletion.
Updating Data
180
FIGURE 6-12
Results of update and delete
Q&A Question: What happens when you run a DELETE command that does not contain a WHERE clause? Answer: Without a condition to specify which row(s) to delete, the query will delete all rows from the table.
Executing a Rollback The following example executes a rollback.
EXAMPLE 7 Execute a rollback and then display the data in the LEVEL1_CUSTOMER table.
To execute a rollback, execute the ROLLBACK command, as shown in Figure 6-13.
ROLLBACK command reverses all changes since last COMMIT
FIGURE 6-13
Executing a rollback
Figure 6-14 shows a SELECT command for the LEVEL1_CUSTOMER table after executing the rollback. Notice that the name of customer 356 has changed back to Ferguson’s and
Chapter 6
the row for customer 895 has been reinstated. All updates made prior to the previous commit are still reflected in the data.
Customer name before update
181 Deleted row reinstated
FIGURE 6-14
Data in the LEVEL1_CUSTOMER table after executing a rollback
ACCESS USER NOTE If you are using Access to complete these steps, you will not be able to execute the ROLLBACK command. Consequently, your data for the remaining examples in this chapter will differ slightly from the data shown in the figures—customer 356 will be named Smith Sport and customer 895 will not be included.
NOTE In the remaining examples in this chapter, the Autocommit feature is enabled, that is, there is a check mark in the Autocommit check box. All updates are committed immediately without requiring any special action on your part. In addition, it will no longer be possible to roll back updates.
CHANGING A VALUE IN A COLUMN TO NULL There are some special issues involved when dealing with nulls. You already have seen how to add a row in which some of the values are null and how to select rows in which a given column is null. You also must be able to change the value in a column in an existing row to null, as shown in Example 8. Remember that to make this type of change, the affected column must accept nulls. If you specified NOT NULL for the column when you created the table, then changing a value in a column to null is prohibited.
EXAMPLE 8 Change the balance of customer 725 in the LEVEL1_CUSTOMER table to null.
Updating Data
The command for changing a value in a column to null is exactly what it would be for changing any other value. You simply use the value NULL as the replacement value, as shown in Figure 6-15. Notice that the value NULL is not enclosed in single quotation marks. If it were, the command would change the balance to the word NULL.
Changes the value to null
182
FIGURE 6-15
Changing a value in a column to null
Figure 6-16 shows the data in the LEVEL1_CUSTOMER table after changing the BALANCE column value for customer 725 to null. In Oracle 10g, a null value is displayed as a hyphen, as shown in Figure 6-16.
Null value
FIGURE 6-16
BALANCE column for customer 725 is null
SQL SERVER USER NOTE In SQL Server, the word “NULL” appears in the results (without the quotation marks) when a column contains a null value.
C H A N G I N G A TA B L E ’S S T R U C T U R E One of the nicest features of a relational DBMS is the ease with which you can change table structures. In addition to adding new tables to the database and deleting tables that are no longer required, you can add new columns to a table and change the physical characteristics of existing columns. Next, you will see how to accomplish these changes. Chapter 6
You can change a table’s structure in SQL by using the ALTER TABLE command, as illustrated in the following examples.
EXAMPLE 9 Premiere Products decides to maintain a customer type for each customer in the database. These types are R for regular customers, D for distributors, and S for special customers. Add this information in a new column named CUSTOMER_TYPE in the LEVEL1_CUSTOMER table.
183
To add a new column, use the ADD clause of the ALTER TABLE command. The format for the ALTER TABLE command is the words ALTER TABLE followed by the name of the table to be altered and an appropriate clause. The ADD clause consists of the word ADD followed by the name of the column to be added, followed by the characteristics of the column. Figure 6-17 shows the appropriate ALTER TABLE command for this example.
Table to alter Column to add
FIGURE 6-17
Adding a column to an existing table
The LEVEL1_CUSTOMER table now contains a column named CUSTOMER_TYPE, a CHAR column with a length of 1. Any new rows added to the table must include values for the new column. Effective immediately, all existing rows also contain this new column. The data in any existing row will contain the new column the next time the row is updated. Any time a row is selected for any reason, however, the system treats the row as though the column is actually present. Thus, to the user, it seems as though the structure was changed immediately. For existing rows, you must assign some value to the CUSTOMER_TYPE column. The simplest approach (from the point of view of the DBMS, not the user) is to assign the value NULL as a CUSTOMER_TYPE in all existing rows. This process requires the CUSTOMER_TYPE column to accept null values, and some systems actually insist on this. The default for Oracle, Access, and SQL Server is to accept null values. To change the values in a new column that was added using an ALTER TABLE command, follow the ALTER TABLE command with an UPDATE command like the one shown in Figure 6-18, which sets the CUSTOMER_TYPE value for all rows to R.
Updating Data
Omitting the WHERE clause updates all rows
184
FIGURE 6-18
Making the same update for all rows
The SELECT command shown in Figure 6-19 verifies that the value in the CUSTOMER_TYPE column for all rows is R.
FIGURE 6-19
CUSTOMER_TYPE set to R for all rows
EXAMPLE 10 Two customers in the LEVEL1_CUSTOMER table have a type other than R. Change the types for customers 842 and 148 to S and D, respectively.
Example 9 used an UPDATE command to assign type R to every customer. To change individual types to something other than type R, use the UPDATE command. Figure 6-20 shows the UPDATE command to change customer 842 to customer type S.
Chapter 6
185
FIGURE 6-20
Updating customer 842 to customer type S
Figure 6-21 shows the UPDATE command to change customer 148 to customer type D.
FIGURE 6-21
Updating customer 148 to customer type D
The SELECT command shown in Figure 6-22 shows the results of these UPDATE commands. The customer type for customer 842 is S and the type for customer 148 is D. The type for all other customers is R.
FIGURE 6-22
Customer types in the LEVEL1_CUSTOMER table after updates
Figure 6-23 uses the DESCRIBE command to display the structure of the LEVEL1_CUSTOMER table, which now includes the CUSTOMER_TYPE column.
Updating Data
DESCRIBE command
New CUSTOMER_TYPE column
186
FIGURE 6-23
Structure of the LEVEL1_CUSTOMER table
ACCESS USER NOTE In Access, use the Documenter tool to show the layout of a table.
SQL SERVER USER NOTE In SQL Server, execute the following command to list all the columns in the LEVEL1_CUSTOMER table: Exec sp_columns LEVEL1_CUSTOMER
EXAMPLE 11 The length of the CUSTOMER_NAME column in the LEVEL1_CUSTOMER table is too short. Increase its length to 50 characters. In addition, change the CREDIT_LIMIT column so it cannot accept nulls.
You can change the characteristics of existing columns by using the MODIFY clause of the ALTER TABLE command. Figure 6-24 shows the ALTER TABLE command that changes the length of the CUSTOMER_NAME column from 35 to 50 characters.
FIGURE 6-24
Chapter 6
Changing the length of the CUSTOMER_NAME column in the LEVEL1_CUSTOMER table
SQL SERVER USER NOTE To change the length of the CUSTOMER_NAME column in SQL Server, use the following ALTER COLUMN clause: ALTER TABLE LEVEL1_CUSTOMER ALTER COLUMN CUSTOMER_NAME CHAR(50);
ACCESS USER NOTE 187
The version of the ALTER TABLE command shown in Figure 6-24 is not available in Access; to modify the table’s structure, make the changes in Design view and save the table.
NOTE You also can decrease the length of columns, but you might lose some data currently in the column. For example, if you decrease the length of the CUSTOMER_NAME column from 35 to 20 characters, only the first 20 characters of the current customer names will be included. Any characters from position 21 on will be lost. Thus, you should only decrease column lengths when you are positive that you will not lose any data stored in the column. You can change the length of DECIMAL columns in the same manner that you change the length of CHAR columns.
Figure 6-25 shows the ALTER TABLE command to change the CREDIT_LIMIT column so it does not accept null values.
FIGURE 6-25 Changing the CREDIT_LIMIT column in the LEVEL1_CUSTOMER table to reject null values
The DESCRIBE command shown in Figure 6-26 shows the revised structure of the LEVEL1_CUSTOMER table. The length of the CUSTOMER_NAME column is 50 characters.
Updating Data
The dash in the Nullable column for the CREDIT_LIMIT column (instead of a check mark) indicates that the CREDIT_LIMIT column no longer accepts null values.
No check mark indicates that the CREDIT_LIMIT column will not accept null values
188
Length changed to 50
FIGURE 6-26
Revised structure of the LEVEL1_CUSTOMER table
NOTE You also can use the MODIFY clause of the ALTER TABLE command to change a column that currently rejects null values so that it accepts null values by using NULL in place of NOT NULL in the ALTER TABLE command.
NOTE If there were existing rows in the LEVEL1_CUSTOMER table in which the CREDIT_LIMIT column was already null, the DBMS would reject the modification to the CREDIT_LIMIT column shown in Figure 6-25 and display an error message indicating that this change is not possible. In this case, you first must use an UPDATE command to change all values that are null to some other value. Then you could alter the table’s structure as shown in the figure.
Making Complex Changes In some cases, you might need to change a table’s structure in ways that are either beyond the capabilities of SQL or that are so complex that it would take longer to make the changes than to re-create the table. Perhaps you need to eliminate multiple columns, rearrange the order of several columns, or combine data from two tables into one. For example, if you try to change a column with a data type of VARCHAR to CHAR, SQL still uses VARCHAR when the table contains other variable-length columns. In these situations, you can use a CREATE TABLE command to describe the new table (which must use a different name than the existing table), and then insert values from the existing table into it using the INSERT command combined with an appropriate SELECT command.
Chapter 6
D RO P P I N G A TA B L E As you learned in Chapter 3, you can delete a table that is no longer needed by executing the DROP TABLE command.
EXAMPLE 12 Delete the LEVEL1_CUSTOMER table because it is no longer needed in the Premiere Products database.
189
The command to delete the table is shown in Figure 6-27.
FIGURE 6-27
DROP TABLE command to delete the LEVEL1_CUSTOMER table
When the command shown in Figure 6-27 is executed, the LEVEL1_CUSTOMER table and all its data are permanently removed from the database.
Updating Data
Chapter Summary ●
To create a new table from an existing table, first create the new table by using the CREATE TABLE command. Then use an INSERT command containing a SELECT command to select the desired data to be included from the existing table.
●
Use the UPDATE command to change existing data in a table.
●
Use the INSERT command to add new rows to a table.
●
Use the DELETE command to delete existing rows from a table.
●
Use the COMMIT command to make updates permanent; use the ROLLBACK command to reverse any updates that have not been committed.
●
To change all values in a column to null, use the SET clause followed by the column name, an equal sign, and the word NULL. To change a specific value in a column to null, use a condition to select the row.
●
To add a column to a table, use the ALTER TABLE command with an ADD clause.
●
To change the characteristics of a column, use the ALTER TABLE command with a MODIFY clause.
●
Use the DROP TABLE command to delete a table and all its data.
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Key Terms ADD clause
MODIFY clause
ALTER TABLE
roll back
Autocommit
ROLLBACK
commit
transaction
COMMIT
UPDATE
DELETE
Review Questions 1.
Which command creates a new table?
2.
Which command and clause adds an individual row to a table?
3.
How do you add data from an existing table to another table?
4.
Which command changes data in a table?
5.
Which command removes rows from a table?
6.
In Oracle and in SQL Server, which command makes updates permanent?
7.
In Oracle and in SQL Server, which command reverses updates? Which updates are reversed?
8.
How do you use the COMMIT and ROLLBACK commands to support transactions?
9.
What is the format of the SET clause that changes the value in a column to null in an UPDATE command?
10. Chapter 6
Which command and clause adds a column to an existing table?
11.
In Oracle and in SQL Server, which command and clause changes the characteristics of an existing column in a table?
12.
Which command deletes a table and all its data?
13.
Microsoft Access supports make-table queries. What is a make-table query? What SQL statement(s) are equivalent to make-table queries?
14.
Use your favorite Web browser and Web search engine to find the SQL command to delete a column in a table. Write the SQL command in Oracle to delete the CUSTOMER_TYPE column from the LEVEL1_CUSTOMER table. Would you use the same command in SQL Server to delete the column? If no, write the command to use in SQL Server.
191
Exercises Premiere Products Use SQL to make the following changes to the Premiere Products database (see Figure 1-2 in Chapter 1). After each change, execute an appropriate query to show that the change was made correctly. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
Create a NONAPPLIANCE table with the structure shown in Figure 6-28.
NONAPPLIANCE Column
Type
Length
PART_NUM
CHAR
4
DESCRIPTION
CHAR
15
ON_HAND
DECIMAL
4
CLASS
CHAR
2
PRICE
DECIMAL
6
FIGURE 6-28
Decimal Places
Nulls Allowed?
Description
No
Part number (primary key) Part description
0
Number of units on hand Item class
2
Unit price
NONAPPLIANCE table layout
2.
Insert into the NONAPPLIANCE table the part number, part description, number of units on hand, item class, and unit price from the PART table for each part that is not in item class AP.
3.
In the NONAPPLIANCE table, change the description of part number AT94 to “Steam Iron.”
4.
In the NONAPPLIANCE table, increase the price of each item in item class SG by three percent. (Hint: Multiply each price by 1.03.)
5.
Add the following part to the NONAPPLIANCE table: part number: TL92; description: Edge Trimmer; number of units on hand: 11; class: HW; and price: 29.95.
6.
Delete every part in the NONAPPLIANCE table for which the class is SG.
7.
In the NONAPPLIANCE table, change the class for part FD21 to null.
8.
Add a column named ON_HAND_VALUE to the NONAPPLIANCE table. The on-hand value is a seven-digit number with two decimal places that represents the product of the number of units on hand and the price. Then set all values of ON_HAND_VALUE to ON_HAND * PRICE.
Updating Data
9. 10.
In the NONAPPLIANCE table, increase the length of the DESCRIPTION column to 30 characters. Remove the NONAPPLIANCE table from the Premiere Products database.
Henry Books
192
Use SQL to make the following changes to the Henry Books database (Figures 1-4 through 1-7 in Chapter 1). After each change, execute an appropriate query to show that the change was made correctly. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
Create a FICTION table with structure shown in Figure 6-29.
FICTION Column
Type
Length
BOOK_CODE
CHAR
4
TITLE
CHAR
40
PUBLISHER_CODE
CHAR
3
PRICE
DECIMAL
4
FIGURE 6-29
Decimal Places
Nulls Allowed?
Description
No
Book code (primary key) Book title Publisher code
2
Book price
FICTION table layout
2.
Insert into the FICTION table the book code, book title, publisher code, and price from the BOOK table for only those books having type FIC.
3.
The publisher with code LB has decreased the price of its fiction books by four percent. Update the prices in the FICTION table accordingly.
4.
Insert a new book into the FICTION table. The book code is 9946, the title is Cannery Row, the publisher is PE, and the price is 11.95.
5.
Delete the book in the FICTION table having the book code 9883.
6.
The price of the book entitled To Kill a Mockingbird has been increased to an unknown amount. Change the value in the FICTION table to reflect this change.
7.
Add to the FICTION table a new character column named BEST_SELLER that is one character in length. Then set the default value for all columns to N.
8.
Change the BEST_SELLER column in the FICTION table to Y for the book entitled Song of Solomon.
9.
Change the length of the TITLE column in the FICTION table to 50 characters.
10.
Change the BEST_SELLER column in the FICTION table to reject nulls.
11.
Delete the FICTION table from the database.
Alexamara Marina Group Use SQL to make the following changes to the Alexamara Marina Group database (Figures 1-8 through 1-12 in Chapter 1). After each change, execute an appropriate query to show that the change was made correctly. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. Chapter 6
1.
Create a LARGE_SLIP table with the structure shown in Figure 6-30. (Hint: If you have trouble creating the primary key, see Figure 3-31 in Chapter 3.)
LARGE_SLIP Column
Type
Length Decimal Places
Nulls Allowed?
Description
MARINA_NUM
CHAR
4
No
Marina number (primary key)
SLIP_NUM
CHAR
4
No
Slip number in the marina (primary key)
RENTAL_FEE
DECIMAL 8
BOAT_NAME
CHAR
50
Name of boat currently in the slip
OWNER_NUM
CHAR
4
Number of boat owner renting the slip
FIGURE 6-30
2
193
Annual rental fee for the slip
LARGE_SLIP table layout
2.
Insert into the LARGE_SLIP table the marina number, slip number, rental fee, boat name, and owner number for those slips whose length is 40 feet.
3.
Alexamara has increased the rental fee of each large slip by $150. Update the rental fees in the LARGE_SLIP table accordingly.
4.
After increasing the rental fee of each large slip by $150 (Exercise 3), Alexamara decides to decrease the rental fee of any slip whose fee is more than $4,000 by one percent. Update the rental fees in the LARGE_SLIP table accordingly.
5.
Insert a new row into the LARGE_SLIP table. The marina number is 1, the slip number is A4, the rental fee is $3,900, the boat name is Bilmore, and the owner number is FE82.
6.
Delete all slips in the LARGE_SLIP table for which the owner number is TR72.
7.
The name of the boat in marina 1 and slip A1 is in the process of being changed to an unknown name. Change the name of this boat in the LARGE_SLIP table to null.
8.
Add to the LARGE_SLIP table a new character column named CHARTER that is one character in length. (This column will indicate whether the boat is available for chartering.) Set the value for the CHARTER column on all rows to N.
9.
Change the CHARTER column in the LARGE_SLIP table to Y for the slip containing the boat named Our Toy.
10.
Change the length of the BOAT_NAME column in the LARGE_SLIP table to 60 characters.
11.
Change the RENTAL_FEE column in the LARGE_SLIP table to reject nulls.
12.
Delete the LARGE_SLIP table from the database.
Updating Data
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CHAPTER
7
DA TABASE A DMIN I STR ATION LEARNING
OBJECTIVES
Objectives ●
Understand, create, and drop views
●
Recognize the benefits of using views
●
Use a view to update data
●
Grant and revoke users’ database privileges
●
Understand the purpose, advantages, and disadvantages of using an index
●
Create, use, and drop an index
●
Understand and obtain information from the system catalog
●
Use integrity constraints to control data entry
INTRODUCTION There are some special issues involved in managing a database. This process, often called database administration, is especially important when more than one person uses the database. In a business organization, a person or an entire group known as the database administrator is charged with managing the database.
In Chapter 6, you learned about one function of the database administrator: changing the structure of a database. In this chapter, you will see how the database administrator can give each user his or her own view of the database. You will use the GRANT and REVOKE commands to assign different database privileges to different users. You will use indexes to improve database performance. You will learn how a
DBMS stores information about the database structure in an object called the system catalog and how to access that information. Finally, you will learn how to specify integrity constraints that establish rules for the data in the database.
CREATING AND USING VIEWS 196
Most DBMSs support the creation of views. A view is a program’s or an individual user’s picture of the database. The existing, permanent tables in a relational database are called base tables. A view is a derived table because the data in it comes from one or more base tables. To the user, a view appears to be an actual table, but it is not. In many cases, a user can examine table data using a view. Because a view usually includes less information than the full database, its use can represent a great simplification. Views also provide a measure of security, because omitting sensitive tables or columns from a view renders them unavailable to anyone accessing the database through the view. To help you understand the concept of a view, suppose that Juan is interested in the part number, part description, units on hand, and unit price of parts in item class HW. He is not interested in any other columns in the PART table, nor is he interested in any rows that correspond to parts in other item classes. Viewing this data would be simpler for Juan if the other rows and columns were not even present. Although you cannot change the structure of the PART table and omit some of its rows just for Juan, you can do the next best thing. You can provide him with a view that consists of only the rows and columns that he needs to access. A view is defined by creating a defining query, which indicates the rows and columns to include in the view. The SQL command (or the defining query) to create the view for Juan is illustrated in Example 1.
EXAMPLE 1 Create a view named HOUSEWARES that consists of the part number, part description, units on hand, and unit price of each part in item class HW.
To create a view, use the CREATE VIEW command, which includes the words CREATE VIEW, followed by the name of the view, the word AS, and then a query. The CREATE VIEW command shown in Figure 7-1 creates a view of the PART table that contains only the specified columns.
Chapter 7
Name of view Defining query
197
FIGURE 7-1 Creating the HOUSEWARES view
ACCESS USER NOTE Access does not support the CREATE VIEW command. To create a view in Access, create a query to define the view, and then save the query object in the database using the view’s name (for example, HOUSEWARES).
Given the current data in the Premiere Products database, the HOUSEWARES view contains the data shown in Figure 7-2.
HOUSEWARES PART_NUM DESCRIPTION ON_HAND
PRICE
AT94
Iron
50
$24.95
DL71
Cordless Drill
21
$129.95
FD21
Stand Mixer
22
$159.95
FIGURE 7-2 HOUSEWARES view
The data does not actually exist in this form, nor will it ever exist in this form. It is tempting to think that when Juan uses this view, the query is executed and produces some sort of temporary table, named HOUSEWARES, that Juan can access, but this is not what actually happens. Instead, the query acts as a sort of “window” into the database, as shown in Figure 7-3. As far as Juan is concerned, the entire database is just the darker shaded portion of the PART table. Juan can see any change that affects the darker portion of the PART table, but he is totally unaware of any other changes that are made in the database.
Database Administration
PART PART_NUM DESCRIPTION
198
ON_HAND CLASS WAREHOUSE
PRICE
3
$24.95
45 SG
2
$794.95
32 AP
1
$165.00
3
$129.95
Gas Range
21 HW 8 AP
2
$495.00
Washer
12 AP
3
$399.99
FD21
Stand Mixer
22 HW
3
$159.95
KL62
Dryer
12 AP
1
$349.95
KT03
Dishwasher
8 AP
3
$595.00
KV29
Treadmill
9 SG
2
$1,390.00
AT94
Iron
50 HW
BV06
Home Gym
CD52
Microwave Oven
DL71
Cordless Drill
DR93 DW11
FIGURE 7-3
Juan’s view of the PART table
When you create a query that involves a view, the DBMS changes the query to one that selects data from the table(s) in the database that created the view. For example, suppose Juan creates the query shown in Figure 7-4.
Data selected from HOUSEWARES view
FIGURE 7-4
Using the HOUSEWARES view
The DBMS does not execute the query in this form. Instead, it merges the query Juan entered with the query that creates the view to form the query that is actually executed. When the DBMS merges the query that creates the view with Juan’s query to select rows for which the ON_HAND value is less than 25, the query that the DBMS actually executes is: SELECT PART_NUM, DESCRIPTION, ON_HAND, PRICE FROM PART WHERE CLASS = 'HW' AND ON_HAND < 25;
In the query that the DBMS executes, the FROM clause lists the PART table rather than the HOUSEWARES view, the SELECT clause lists columns from the PART table instead of * to select all columns from the HOUSEWARES view, and the WHERE clause contains a compound condition to select only those parts in the HW class (as Juan sees in the Chapter 7
HOUSEWARES view) and only those parts with ON_HAND values of less than 25. This new query is the one that the DBMS actually executes. Juan, however, is unaware that this activity is taking place. To Juan, it seems that he is really using a table named HOUSEWARES. One advantage of this approach is that because the HOUSEWARES view never exists in its own right, any update to the PART table is immediately available in the HOUSEWARES view. If the HOUSEWARES view were really a table, this immediate update would not be possible. You also can assign column names that are different from those in the base table, as illustrated in the next example. 199
EXAMPLE 2 Create a view named HSEWRES that consists of the part number, part description, units on hand, and unit price of all parts in item class HW. In this view, change the names of the PART_NUM, DESCRIPTION, ON_HAND, and PRICE columns to PNUM, DSC, OH, and PRCE, respectively.
When renaming columns, you include the new column names in parentheses following the name of the view, as shown in Figure 7-5. In this case, anyone accessing the HSEWRES view will refer to PART_NUM as PNUM, to DESCRIPTION as DSC, to ON_HAND as OH, and to PRICE as PRCE.
New names for columns listed in parentheses
FIGURE 7-5
Renaming columns when creating a view
If you select all columns from the HSEWRES view, the output displays the new column names, as shown in Figure 7-6.
Database Administration
New column names
200
FIGURE 7-6
Data in the HSEWRES view
ACCESS USER NOTE To change column names in Access, use AS clauses in the query design (for example, SELECT PART_NUM AS PNUM, DESCRIPTION AS DSC, and so on).
The HSEWRES view is an example of a row-and-column subset view because it consists of a subset of the rows and columns in some base table—in this case, in the PART table. Because the defining query can be any valid SQL query, a view also can join two or more tables or involve statistics. The next example illustrates a view that joins two tables.
EXAMPLE 3 Create a view named REP_CUST consisting of the sales rep number (named RNUM), sales rep last name (named RLAST), sales rep first name (named RFIRST), customer number (named CNUM), and customer name (named CNAME) for all sales reps and matching customers in the REP and CUSTOMER tables.
The command to create this view appears in Figure 7-7.
View joins two tables
FIGURE 7-7
Chapter 7
Creating the REP_CUST view
Given the current data in the Premiere Products database, the REP_CUST view contains the data shown in Figure 7-8.
201
FIGURE 7-8 Data in the REP_CUST view
SQL SERVER USER NOTE SQL Server does not support the ORDER BY clause in a CREATE VIEW command. When you need to order the query results, insert an ORDER BY clause in the SELECT command when you query the view. For example, the following SELECT command retrieves all records in the REP_CUST view ordered by rep number and customer number: SELECT * FROM REP_CUST ORDER BY RNUM, CNUM
A view also can involve statistics, as illustrated in Example 4.
EXAMPLE 4 Create a view named CRED_CUST that consists of each credit limit (CREDIT_LIMIT) and the number of customers having this credit limit (NUM_CUSTOMERS). Sort the credit limits in ascending order.
The command shown in Figure 7-9 creates this view.
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FIGURE 7-9
Creating the CRED_CUST view
The SELECT command shown in Figure 7-10 displays the current data in the Premiere Products database for this view.
FIGURE 7-10
Data in the CRED_CUST view
The use of views provides several benefits. First, views provide data independence. When the database structure changes (by adding columns or changing the way objects are related, for example) in such a way that the view still can be derived from existing data, the user can access and use the same view. If adding extra columns to tables in the database is the only change, and these columns are not required by the view’s user, the defining query might not even need to be changed for the user to continue using the view. If table relationships are changed, the defining query might be different, but because users are not aware of the defining query, they are unaware of this difference. Users continue accessing the database through the same view, as though nothing has changed. For example, suppose customers are assigned to territories, each territory is assigned to a single sales rep, a sales rep can have more than one territory, and a customer is represented by the sales rep
Chapter 7
who covers the customer’s assigned territory. To implement these changes, you might choose to restructure the database as follows: REP(REP_NUM, LAST_NAME, FIRST_NAME, STREET, CITY, STATE, ZIP, COMMISSION, RATE) TERRITORY(TERRITORY_NUM, DESCRIPTION, REP_NUM) CUSTOMER(CUSTOMER_NUM, CUSTOMER_NAME, STREET, CITY, STATE, ZIP, BALANCE, CREDIT_LIMIT, TERRITORY_NUM)
Assuming that the REP_CUST view created in Figure 7-7 is still required, you could change the defining query as follows: CREATE VIEW REP_CUST (RNUM, RLAST, RFIRST, CNUM, CNAME) AS SELECT REP.REP_NUM, REP.LAST_NAME, REP.FIRST_NAME, CUSTOMER_NUM, CUSTOMER_NAME FROM REP, TERRITORY, CUSTOMER WHERE REP.REP_NUM = TERRITORY.REP_NUM AND TERRITORY.TERRITORY_NUM = CUSTOMER.TERRITORY_NUM;
203
This view’s user still can retrieve the number and name of a sales rep together with the number and name of each customer the sales rep represents. The user is unaware, however, of the new structure in the database. The second benefit of using views is that different users can see the same data in different ways through their own views. In other words, you can customize the display of data to meet each user’s needs. The final benefit of using views is that a view can contain only those columns required by a given user. This practice has two advantages. First, because the view usually contains fewer columns than the overall database and is conceptually a single table, rather than a collection of tables, a view greatly simplifies the user’s perception of the database. Second, views provide a measure of security. Columns that are not included in the view are not accessible to the view’s user. For example, omitting the BALANCE column from a view ensures that the view’s user cannot access any customer’s balance. Likewise, rows that are not included in the view are not accessible. A user of the HOUSEWARES view, for example, cannot obtain any information about parts in the AP or SG classes.
USING A VIEW TO UPDATE DATA The benefits of using views hold true only when views are used for retrieval purposes. When updating the database, the issues involved in updating data through a view depend on the type of view, as you will see next.
Updating Row-and-Column Subset Views Consider the row-and-column subset view for the HOUSEWARES view. There are columns in the underlying base table (PART) that are not present in the view. If you attempt to add a row with the data ('BB99','PAN',50,14.95), the DBMS must determine how to enter the data in those columns from the PART table that are not included in the HOUSEWARES view (CLASS and WAREHOUSE). In this case, it is clear what data to enter in the CLASS column. According to the view definition, all rows are item class HW, but it is not clear what
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data to enter in the WAREHOUSE column. The only possibility would be NULL. Therefore, if every column not included in a view can accept nulls, you can add new rows using the INSERT command. There is another problem, however. Suppose the user attempts to add a row to the HOUSEWARES view containing the data ('BV06','Waffle Maker',5,29.95). Because part number BV06 already exists in the PART table, the system must reject this attempt. Because this part is not in item class HW (and therefore is not in the HOUSEWARES view), this rejection certainly will seem strange to the user, because there is no such part in the user’s view. On the other hand, updates or deletions cause no particular problem in this view. If the description of part number FD21 changes from Stand Mixer to Pan, this change is made in the PART table. If part number DL71 is deleted, this deletion occurs in the PART table. One surprising change could take place, however. Suppose that the CLASS column is included in the HOUSEWARES view and a user changes the class of part number AT94 from HW to AP. Because this item would no longer satisfy the criterion for being included in the HOUSEWARES view, part number AT94 would disappear from the user’s view! Although there are problems to overcome when updating row-and-column subset views, it seems possible to update the database through the HOUSEWARES view. This does not mean that any row-and-column subset view is updatable, however. Consider the REP_CRED view shown in Figure 7-11. (The DISTINCT operator is used to omit duplicate rows from the view.)
FIGURE 7-11
Creating the REP_CRED view
Figure 7-12 shows the data in the REP_CRED view.
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205
FIGURE 7-12
Data in the REP_CRED view
How would you add the row 15000,'35' to this view? In the underlying base table (CUSTOMER), at least one customer must be added whose credit limit is $15,000 and whose sales rep number is 35, but which customer is it? You cannot leave the other columns null in this case, because one of them is CUSTOMER_NUM, which is the base table’s primary key. What would it mean to change the row 5000,'35' to 15000,'35'? Would it mean changing the credit limit to $15,000 for each customer represented by sales rep number 35 that currently has a credit limit of $5,000? Would it mean changing the credit limit of one of these customers and deleting the rest? What would it mean to delete the row 5000,'35'? Would it mean deleting all customers with credit limits of $5,000 and represented by sales rep number 35, or would it mean assigning these customers a different sales rep or a different credit limit? Why does the REP_CRED view involve a number of serious problems that are not present in the HOUSEWARES view? The basic reason is that the HOUSEWARES view includes, as one of its columns, the primary key of the underlying base table, but the REP_CRED view does not. A row-and-column subset view that contains the primary key of the underlying base table is updatable (subject, of course, to some of the concerns already discussed).
Updating Views Involving Joins In general, views that involve joins of base tables can cause problems when updating data. Consider the relatively simple REP_CUST view, for example, described earlier (see Figures 7-7 and 7-8). The fact that some columns in the underlying base tables are not included in this view presents some of the same problems discussed earlier. Assuming that you can overcome these problems by using nulls, there are more serious problems when attempting to update the database through this view. On the surface, changing the row ('35','Hull','Richard', '282','Brookings Direct') to ('35','Baldwin','Sara','282','Brookings Direct'), might not appear to pose any problems other than some inconsistency in the data. (In the new version of the row,
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the name of sales rep 35 is Sara Baldwin; whereas in the fourth row in the table, the name of sales rep 35, the same sales rep, is Richard Hull.) The problem is actually more serious than that—making this change is not possible. The name of a sales rep is stored only once in the underlying REP table. Changing the name of sales rep 35 from Richard Hull to Sara Baldwin in this one row of the view causes the change to be made to the single row for sales rep 35 in the REP table. Because the view simply displays data from the base tables, for each row on which the sales rep number is 35, the sales rep name is now Sara Baldwin. In other words, it appears that the same change has been made in the other rows. In this case, this change ensures consistency in the data. In general, however, the unexpected changes caused by an update are not desirable. Before concluding the topic of views that involve joins, you should note that all joins do not create the preceding problem. When two base tables have the same primary key and the primary key is used as the join column, updating the database using the view is not a problem. For example, suppose the actual database contains two tables (REP_DEMO and REP_FIN) instead of one table (REP). Figure 7-13 shows the data in the REP_DEMO table.
FIGURE 7-13
Data in the REP_DEMO table
Figure 7-14 shows the data in the REP_FIN table.
FIGURE 7-14
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Data in the REP_FIN table
What was once a single table in the original Premiere Products design has been divided into two separate tables. Users who need to see the rep data in a single table can use a view named SALES_REP that joins these two tables using the REP_NUM column. The defining query for the SALES_REP view appears in Figure 7-15.
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View joins REP_DEMO and REP_FIN tables
FIGURE 7-15
Creating the SALES_REP view
The data in the SALES_REP view appears in Figure 7-16.
FIGURE 7-16
Data in the SALES_REP view
It is easy to update the SALES_REP view. To add a row, use an INSERT command to add a row to each underlying base table. To update data in a row, make the change in the appropriate base table. To delete a row from the view, delete the corresponding rows from both underlying base tables.
Q&A Question: How would you add the row ('10','Peters','Jean','14 Brink','Holt','FL','46223', 107.50,0.05) to the SALES_REP view? Answer: Use an INSERT command to add the row ('10','Peters','Jean','14 Brink','Holt','FL','46223') to the REP_DEMO table, and then use another INSERT command to add the row ('10',107.50,0.05) to the REP_FIN table.
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Q&A Question: How would you change the name of sales rep 20 to Valerie Lewis? Answer: Use an UPDATE command to change the name in the REP_DEMO table.
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Question: How would you change Valerie’s commission rate to 0.06? Answer: Use an UPDATE command to change the rate in the REP_FIN table.
Q&A Question: How would you delete sales rep 35 from the REP table? Answer: Use a DELETE command to delete sales rep 35 from both the REP_DEMO and REP_FIN tables.
Updates (additions, changes, or deletions) to the SALES_REP view do not cause any problems. The main reason that the SALES_REP view is updatable—and that other views involving joins might not be updatable—is that this view is derived from joining two base tables on the primary key of each table. In contrast, the REP_CUST view is created by joining two tables by matching the primary key of one table with a column that is not the primary key in the other table. When neither of the join columns in a view is a primary key column, users will encounter even more severe problems when attempting to make updates.
Updating Views Involving Statistics A view that involves statistics calculated from one or more base tables is the most troublesome view when attempting to update data. Consider the CRED_CUST view, for example (see Figure 7-10). How would you add the row 9000,3 to indicate that there are three customers that have credit limits of $9,000 each? Likewise, changing the row 5000,2 to 5000,5 means you are adding three new customers with credit limits of $5,000 each, for a total of five customers. Clearly these are impossible tasks; you cannot add rows to a view that includes calculations.
DROPPING A VIEW When a view is no longer needed, you can remove it using the DROP VIEW command.
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EXAMPLE 5 The HSEWRES view is no longer necessary, so delete it.
The command to delete a view is DROP VIEW as shown in Figure 7-17.
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FIGURE 7-17
Dropping a view
ACCESS USER NOTE Access does not support the DROP VIEW command. To drop a view, delete the query object you saved when you created the view.
SECURITY Security is the prevention of unauthorized access to a database. Within an organization, the database administrator determines the types of access various users need to the database. Some users might need to retrieve and update anything in the database. Other users might need to retrieve any data from the database but not make any changes to it. Still other users might need to access only a portion of the database. For example, Bill might need to retrieve and update customer data, but does not need to access data about sales reps, orders, order lines, or parts. Valerie might need to retrieve part data and nothing else. Sam might need to retrieve and update data on parts in the HW class, but does not need to retrieve data in any other classes. After the database administrator has determined the access different users of the database need, the DBMS enforces these access rules by whatever security mechanism the DBMS supports. You can use SQL to enforce two security mechanisms. You already have seen that views furnish a certain amount of security; when users are accessing the database through a view, they cannot access any data that is not included in the view. The main mechanism for providing access to a database, however, is the GRANT command. The basic idea of the GRANT command is that the database administrator can grant different types of privileges to users and then revoke them later, if necessary. These privileges include the right to select, insert, update, and delete table data. You can grant and revoke user privileges using the GRANT and REVOKE commands. The following examples illustrate various uses of the GRANT command when the named users already exist in the database.
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NOTE Do not execute the commands in this section unless your instructor asks you to do so.
EXAMPLE 6 User Johnson must be able to retrieve data from the REP table.
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The following GRANT command permits a user named Johnson to execute SELECT commands for the REP table: GRANT SELECT ON REP TO JOHNSON;
EXAMPLE 7 Users Smith and Brown must be able to add new parts to the PART table.
The following GRANT command permits two users named Smith and Brown to execute INSERT commands for the PART table. Notice that a comma separates the user names: GRANT INSERT ON PART TO SMITH, BROWN;
EXAMPLE 8 User Anderson must be able to change the name and street address of customers.
The following GRANT command permits a user named Anderson to execute UPDATE commands involving the CUSTOMER_NAME and STREET columns in the CUSTOMER table. Notice that the SQL command includes the column names in parentheses before the ON clause: GRANT UPDATE (CUSTOMER_NAME, STREET) ON CUSTOMER TO ANDERSON;
EXAMPLE 9 User Thompson must be able to delete order lines.
The following GRANT command permits a user named Thompson to execute DELETE commands for the ORDER_LINE table: GRANT DELETE ON ORDER_LINE TO THOMPSON;
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EXAMPLE 10 Every user must be able to retrieve part numbers, part descriptions, and item classes.
The GRANT command to indicate that all users can retrieve data using a SELECT command includes the word PUBLIC, as follows: GRANT SELECT (PART_NUM, DESCRIPTION, CLASS) ON PART TO PUBLIC;
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EXAMPLE 11 User Roberts must be able to create an index on the REP table.
You will learn about indexes and their uses in the next section. The following GRANT command permits a user named Roberts to create an index on the REP table: GRANT INDEX ON REP TO ROBERTS;
EXAMPLE 12 User Thomas must be able to change the structure of the CUSTOMER table.
The following GRANT command permits a user named Thomas to execute ALTER commands for the CUSTOMER table so he can change the table’s structure: GRANT ALTER ON CUSTOMER TO THOMAS;
EXAMPLE 13 User Wilson must have all privileges for the REP table.
The GRANT command to indicate that a user has all privileges includes the ALL privilege, as follows: GRANT ALL ON REP TO WILSON;
The privileges that a database administrator can grant are SELECT to retrieve data, UPDATE to change data, DELETE to delete data, INSERT to add new data, INDEX to create an index, and ALTER to change the table structure. The database administrator usually assigns privileges. Normally, when the database administrator grants a particular privilege to a user, the user cannot pass that privilege along to other users. When the user needs to be able to pass the privilege to other users, the GRANT command must include the Database Administration
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WITH GRANT OPTION clause. This clause grants the indicated privilege to the user and also permits the user to grant the same privileges (or a subset of them) to other users. The database administrator uses the REVOKE command to revoke privileges from users. The format of the REVOKE command is essentially the same as that of the GRANT command, but with two differences: the word GRANT is replaced by the word REVOKE, and the word TO is replaced by the word FROM. In addition, the clause WITH GRANT OPTION obviously is not meaningful as part of a REVOKE command. Incidentally, the revoke cascades, so if Johnson is granted privileges WITH GRANT OPTION and then Johnson grants these same privileges to Smith, revoking the privileges from Johnson revokes Smith’s privileges at the same time. Example 14 illustrates the use of the REVOKE command.
EXAMPLE 14 User Johnson is no longer allowed to retrieve data from the REP table.
The following REVOKE command revokes the SELECT privilege for the REP table from the user named Johnson: REVOKE SELECT ON REP FROM JOHNSON;
The database administrator can also apply the GRANT and REVOKE commands to views to restrict access to only certain rows within tables.
INDEXES When you query a database, you are usually searching for a row (or collection of rows) that satisfies some condition. Examining every row in a table to find the ones you need often takes too much time to be practical, especially in tables with thousands of rows. Fortunately, you can create and use an index to speed up the searching process significantly. An index in SQL is similar to an index in a book. When you want to find a discussion of a given topic in a book, you could scan the entire book from start to finish, looking for references to the topic you need. More than likely, however, you would not have to resort to this technique. If the book has a good index, you could use it to identify the pages on which your topic is discussed. In a DBMS, the main mechanism for increasing the efficiency with which data is retrieved from the database is the index. Conceptually, these indexes are very much like the index in a book. Consider Figure 7-18, for example, which shows the CUSTOMER table for Premiere Products together with one extra column named ROW_NUMBER. This extra column gives the location of the row in the table (customer 148 is the first row in the table and is on row 1, customer 282 is on row 2, and so on). The DBMS—not the user—automatically assigns and uses these row numbers, and that is why you do not see them.
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CUSTOMER ROW_ CUSTOMER_ CUSTOMER_ NAME NUMBER NUM
STREET
CITY
1
148
Al’s Appliance and Sport
2
282
Brookings Direct
3
356
Ferguson’s
4
408
5
462
The Everything 1828 Shop Raven Bargains Galore 3829 Central
6
524
Kline’s
7
608
Johnson’s Department Store
8
687
9
725
10
842
Lee’s Sport and 282 Altonville Evergreen Appliance Sheldon Deerfield’s Four 282 Columbia Seasons 28 All Season Grove Lakeview
FIGURE 7-18
STATE ZIP
BALANCE
CREDIT_ REP_ LIMIT NUM
FL
33336
FL
33321
Northfield FL
33146
$5,785.00
$7,500.00 65
Crystal
FL
33503
$5,285.25
$5,000.00 35
Grove
FL
33321
$3,412.00 $10,000.00 65
838 Ridgeland
Fillmore
FL
33336 $12,762.00 $15,000.00 20
372 Oxford
Sheldon
FL
33553
$2,106.00 $10,000.00 65
FL
32543
$2,851.00
$5,000.00 35
FL
33553
$248.00
$7,500.00 35
FL
33321
$8,221.00
$7,500.00 20
2837 Fillmore Greenway 3827 Grove Devon 382 Wildwood
$6,550.00
$7,500.00 20
$431.50 $10,000.00 35
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CUSTOMER table with row numbers
To access a customer’s row using its customer number, you might create and use an index, as shown in Figure 7-19. The index has two columns: the first column contains a customer number, and the second column contains the number of the row on which the customer number is found. To find a customer, look up the customer’s number in the first column in the index. The value in the second column indicates which row to retrieve from the CUSTOMER table, then the row for the desired customer is retrieved.
CUSTOMER_NUM Index CUSTOMER_NUM ROW_NUMBER 148
1
282
2
356
3
408
4
462
5
524
6
608
7
687
8
725
9
842
10
FIGURE 7-19
Index for the CUSTOMER table on the CUSTOMER_NUM column
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Because customer numbers are unique, there is only a single row number in this index. This is not always the case, however. Suppose you need to access all customers with a specific credit limit or all customers represented by a specific sales rep. You might choose to create and use an index on the CREDIT_LIMIT column and an index on the REP_NUM column, as shown in Figure 7-20. In the CREDIT_LIMIT index, the first column contains a credit limit and the second column contains the numbers of all rows on which that credit limit appears. The REP_NUM index is similar, except that the first column contains a sales rep number. 214
CREDIT_LIMIT Index CREDIT_LIMIT ROW_NUMBER
REP_NUM Index REP_NUM ROW_NUMBER 1, 6, 10
$5,000.00
4, 8
20
$7,500.00
1, 3, 9, 10
35
2, 4, 8, 9
65
3, 5, 7
$10,000.00
2, 5, 7
$15,000.00
6
FIGURE 7-20
Indexes for the CUSTOMER table on the CREDIT_LIMIT and REP_NUM columns
Q&A Question: How would you use the index shown in Figure 7-20 to find every customer with a $10,000 credit limit? Answer: Look up $10,000 in the CREDIT_LIMIT index to find a collection of row numbers (2, 5, and 7). Use these row numbers to find the corresponding rows in the CUSTOMER table (Brookings Direct, Bargains Galore, and Johnson’s Department Store).
Q&A Question: How would you use the index shown in Figure 7-20 to find every customer represented by sales rep 35? Answer: Look up 35 in the REP_NUM index to find a collection of row numbers (2, 4, 8, and 9). Use these row numbers to find the corresponding rows in the CUSTOMER table (Brookings Direct, The Everything Shop, Lee’s Sport and Appliance, and Deerfield’s Four Seasons).
The actual structure of an index is more complicated than what is shown in the figures. Fortunately, you do not have to worry about the details of manipulating and using indexes because the DBMS manages them for you—your only job is to determine the columns on which to build the indexes. Typically, you can create and maintain an index for any column or combination of columns in any table. After creating an index, the DBMS uses it to speed up data retrieval.
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As you would expect, the use of any index has advantages and disadvantages. An important advantage was already mentioned: an index makes certain types of retrieval more efficient. There are two disadvantages when using indexes. First, an index occupies disk space. Using this space for an index, however, is technically unnecessary because any retrieval that you can make using an index also can be made without the index; the index just speeds up the retrieval. The second disadvantage is that the DBMS must update the index whenever corresponding data in the database is updated. Without the index, the DBMS would not need to make these updates. The main question that you must ask when considering whether to create a given index is this: do the benefits derived during retrieval outweigh the additional storage required and the extra processing involved in update operations? In a very large database, you might find that indexes are essential to decrease the time required to retrieve records. In a small database, however, an index might not provide any significant benefits. You can add and drop indexes as necessary. You can create an index after the database is built; it does not need to be created at the same time as the database. Likewise, when an existing index is no longer necessary, you can drop it.
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Creating an Index Suppose some users at Premiere Products need to display customer records ordered by balance. Other users need to access a customer’s name using the customer’s number. In addition, some users need to produce a report in which customer records are listed by credit limit in descending order. Within the group of customers having the same credit limit, the customer records must be ordered by name. Each of the previous requirements is carried out more efficiently when you create the appropriate index. The command used to create an index is CREATE INDEX, as illustrated in Example 15.
EXAMPLE 15 Create an index named BALIND on the BALANCE column in the CUSTOMER table. Create an index named REPNAME on the combination of the LAST_NAME and FIRST_NAME columns in the REP table. Create an index named CREDNAME on the combination of the CREDIT_LIMIT and CUSTOMER_NAME columns in the CUSTOMER table, with the credit limits listed in descending order.
The CREATE INDEX command to create the index named BALIND appears in Figure 7-21. The command lists the name of the index and the table name on which the index is to be created. The column on which to create the index—BALANCE—is listed in parentheses.
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FIGURE 7-21 216
Creating the BALIND index on the BALANCE column
The CREATE INDEX command to create the index named REPNAME on the combination of the LAST_NAME and FIRST_NAME columns in the REP table appears in Figure 7-22.
FIGURE 7-22
Creating the REPNAME index on the LAST_NAME and FIRST_NAME columns
The CREATE INDEX command to create the index named CREDNAME on the combination of the CREDIT_LIMIT and CUSTOMER_NAME columns in the CUSTOMER table appears in Figure 7-23. When you need to index a column in descending order, the column name is followed by the DESC operator.
Descending order
FIGURE 7-23
Creating the CREDNAME index on the CREDIT_LIMIT and CUSTOMER_NAME columns
When customers are listed using the CREDNAME index, the records appear in order by descending credit limit. Within any credit limit, the customers are listed alphabetically by name.
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ACCESS USER NOTE Access supports the creation of indexes in both SQL view and Design view.You can write a CREATE INDEX command in SQL view. You can also open the table containing the column(s) on which you want to create an index in Design view, and click the Indexes button or use the Indexed property to create the index.
Dropping an Index The command used to drop (delete) an index is DROP INDEX, which consists of the words DROP INDEX followed by the name of the index to drop. To delete the CREDNAME index on the CUSTOMER table, for example, the command is:
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DROP INDEX CREDNAME;
The DROP INDEX command permanently deletes the index. CREDNAME was the index the DBMS used when listing customer records in descending order by credit limit order and then by customer name within credit limit. The DBMS still can list customers in this order; however, it cannot do so as efficiently without the index.
SQL SERVER USER NOTE The SQL Server command to drop an index requires that you qualify the index name. To delete the CREDNAME index on the CUSTOMER table, for example, the command is: DROP INDEX CUSTOMER.CREDNAME;
Creating Unique Indexes When you specify a table’s primary key, the DBMS automatically ensures that the values entered in the primary key column(s) are unique. For example, the DBMS rejects an attempt to add a second customer whose number is 148 in the CUSTOMER table because customer 148 already exists. Thus, you do not need to take any special action to make sure that values in the primary key column are unique; the DBMS does it for you. Occasionally, a nonprimary key column might store unique values. For example, in the REP table, the primary key is REP_NUM. If the REP table also contains a column for Social Security numbers, the values in this column also must be unique because no two people can have the same Social Security number. Because the Social Security number column is not the table’s primary key, however, you need to take special action in order for the DBMS to ensure that there are no duplicate values in this column. To ensure the uniqueness of values in a nonprimary key column, you can create a unique index by using the CREATE UNIQUE INDEX command. To create a unique index named SSN on the SOC_SEC_NUM column in the REP table, for example, the command is: CREATE UNIQUE INDEX SSN ON REP(SOC_SEC_NUM);
This unique index has all the properties of indexes already discussed, along with one additional property: the DBMS rejects any update that causes a duplicate value in the SOC_SEC_NUM column. In this case, the DBMS rejects the addition of a rep whose Social Security number is the same as that of another rep already in the database.
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SYSTEM CATALOG
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Information about the tables in the database is kept in the system catalog (catalog) or the data dictionary. This section describes the types of items kept in the catalog and the way in which you can query it to access information about the database structure. The DBMS automatically maintains the system catalog, which contains several tables. The catalog tables you’ll consider in this basic introduction are SYSTABLES (information about the tables known to SQL), SYSCOLUMNS (information about the columns within these tables), and SYSVIEWS (information about the views that have been created). Individual SQL implementations might use different names for these tables. In Oracle, the equivalent tables are named DBA_TABLES, DBA_TAB_COLUMNS, and DBA_VIEWS. The system catalog is a relational database of its own. Consequently, you can use the same types of queries to retrieve information that you can use to retrieve data in a relational database. You can obtain information about the tables in a relational database, the columns they contain, and the views built on them from the system catalog. The following examples illustrate this process.
NOTE Most Oracle users need privileges to view system catalog data, so you might not be able to execute these commands. If you are executing the commands shown in the figures, substitute your user name for PRATT to list objects that you own. Your results will differ from those shown in the figures.
ACCESS USER NOTE In Access, use the Documenter to obtain the information discussed in this section, rather than querying the system catalog.
SQL SERVER USER NOTE In SQL Server, use stored procedures to obtain the information discussed in this section. To display information about the tables and views in a database, use the sp_tables procedure. For example, the following command displays all the tables and views in the current database: EXEC sp_tables The sp_columns stored procedure displays information about the columns in a particular table. The following command displays the column information for the REP table: EXEC sp_columns REP
EXAMPLE 16 List the name of each table for which the owner (creator of the table) is PRATT.
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The command to list the table names owned by PRATT is shown in Figure 7-24. The WHERE clause restricts the tables to only those owned by PRATT.
DBA_TABLES is sometimes called SYSTABLES
Table owner must be PRATT
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FIGURE 7-24 Tables owned by PRATT
EXAMPLE 17 List the name of each view owned by PRATT.
This command is similar to the command in Example 16. Rather than TABLE_NAME, the column to be selected is named VIEW_NAME. The command appears in Figure 7-25.
DBA_VIEWS is sometimes called SYSVIEWS
FIGURE 7-25
Views owned by PRATT
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EXAMPLE 18 For the CUSTOMER table owned by PRATT, list each column and its data type.
The command for this example appears in Figure 7-26. The columns to select are COLUMN_NAME and DATA_TYPE.
220 DBA_TAB_COLUMNS is sometimes called SYSCOLUMNS
FIGURE 7-26
Columns in the CUSTOMER table
EXAMPLE 19 List each table owned by PRATT that contains a column named CUSTOMER_NUM.
As shown in Figure 7-27, the COLUMN_NAME column is used in the WHERE clause to restrict the rows to those in which the column name is CUSTOMER_NUM. (Extra tables generated by Oracle might appear in your list, as shown in Figure 7-27.)
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FIGURE 7-27
Tables owned by PRATT with CUSTOMER_NUM columns
When users create, alter, or drop tables or create or drop indexes, the DBMS updates the system catalog automatically to reflect these changes. Users should not execute SQL queries to update the catalog directly because this might produce inconsistent information. For example, when a user deletes the CUSTOMER_NUM column in the DBA_TAB_COLUMNS table, the DBMS would no longer have any knowledge of this column, which is the CUSTOMER table’s primary key, yet all the rows in the CUSTOMER table would still contain a customer number. The DBMS might now treat those customer numbers as names, because as far as the DBMS is concerned, the column named CUSTOMER_NAME is the first column in the CUSTOMER table.
INTEGRITY CONSTRAINTS IN SQL An integrity constraint is a rule for the data in the database. Examples of integrity constraints in the Premiere Products database are as follows: ● ●
●
A sales rep’s number must be unique. The sales rep number for a customer must match the number of a sales rep currently in the database. For example, because there is no sales rep number 11, a customer cannot be assigned to sales rep 11. Item classes for parts must be AP, HW, or SG because these are the only valid item classes.
If a user enters data in the database that violates any of these integrity constraints, the database develops serious problems. For example, two sales reps with the same number, a customer with a nonexistent sales rep, or a part in a nonexistent item class would compromise the integrity of data in the database. To manage these types of problems, SQL provides integrity support, the process of specifying and enforcing integrity constraints for a database. SQL has clauses to support three types of integrity constraints that you can specify within a CREATE TABLE or an ALTER TABLE command. The only difference between these two commands is that an ALTER TABLE command is followed by the word
Database Administration
ADD to indicate that you are adding the constraint to the list of existing constraints. To change an integrity constraint after it has been created, just enter the new constraint, which immediately takes the place of the original. The types of constraints supported in SQL are primary keys, foreign keys, and legal values. In most cases, you specify a table’s primary key when you create the table. To add a primary key after creating a table, you can use the ADD PRIMARY KEY clause of the ALTER TABLE command. For example, to indicate that REP_NUM is the primary key for the REP table, the ALTER TABLE command is: 222
ALTER TABLE REP ADD PRIMARY KEY (REP_NUM);
The ADD PRIMARY KEY clause is ADD PRIMARY KEY followed by the column name that makes up the primary key in parentheses. When the primary key contains more than one column, use commas to separate the column names.
ACCESS USER NOTE To specify a table’s primary key in Access, open the table in Design view, select the column(s) that make up the primary key, and then click the Primary Key button on the Ribbon or the toolbar.
A foreign key is a column in one table whose values match the primary key in another table. (One example is the CUSTOMER_NUM column in the ORDERS table. Values in this column are required to match those of the primary key in the CUSTOMER table.)
EXAMPLE 20 Specify the CUSTOMER_NUM column in the ORDERS table as a foreign key that must match the CUSTOMER table.
When a table contains a foreign key, you identify it using the ADD FOREIGN KEY clause of the ALTER TABLE command. In this clause, you specify the column that is a foreign key and the table it matches. The general form for assigning a foreign key is ADD FOREIGN KEY, the column name(s) of the foreign key, the REFERENCES clause, and then the table name that the foreign key must match, as shown in Figure 7-28.
Name of associated table Column name
FIGURE 7-28
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Adding a foreign key to an existing table
ACCESS USER NOTE To specify a foreign key in Access, open the Relationships window, relate the corresponding tables on the matching column, and then select the option to enforce referential integrity.
After creating a foreign key, the DBMS rejects any update that violates the foreign key constraint. For example, the DBMS rejects the INSERT command shown in Figure 7-29 because it attempts to add an order for which the customer number (850) does not match any customer in the CUSTOMER table. 223 invalid customer number
Error message
FIGURE 7-29 Violating a foreign key constraint when adding a row
The DBMS also rejects the DELETE command in Figure 7-30 because it attempts to delete customer number 148; rows in the ORDERS table for which the customer number is 148 would no longer match any row in the CUSTOMER table.
Command to delete a customer with orders
Error message
FIGURE 7-30
Violating a foreign key constraint when deleting a row
Note that the error messages shown in Figures 7-29 and 7-30 include the words “parent” and “child.” When you specify a foreign key, the table containing the foreign key is the child, and the table referenced by the foreign key is the parent. For example, the CUSTOMER_NUM column in the ORDERS table is a foreign key that references the CUSTOMER table. For this foreign key, the CUSTOMER table is the parent, and the ORDERS table is the child. The error message shown in Figure 7-29 indicates that there is no parent for the order (there is no customer number 850). The error message shown in Figure 7-30 indicates that there are child records (rows) for customer 148 (customer 148 has orders). The DBMS rejects both updates because they violate referential integrity. Database Administration
EXAMPLE 21 Specify the valid item classes for the PART table as AP, HW, and SG.
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You use the CHECK clause of the ALTER TABLE command to ensure that only legal values satisfying a particular condition are allowed in a given column. The general form of the CHECK clause is the word CHECK followed by a condition. If a user enters data that violates the condition, the DBMS rejects the update automatically. For example, to ensure that the only legal values for item class are AP, HW, or SG, use one of the following versions of the CHECK clause: CHECK (CLASS IN ('AP', 'HW', 'SG') )
or CHECK (CLASS = 'AP' OR CLASS = 'HW' OR CLASS = 'SG')
The ALTER TABLE command shown in Figure 7-31 uses the first version of the CHECK clause.
Valid classes are AP, HW, and SG
FIGURE 7-31
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Adding an integrity constraint to an existing table
Now the DBMS will reject the update shown in Figure 7-32 because the command attempts to change the item class to XX, which is an illegal value.
Command to change a part’s class to an invalid value
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Error message
FIGURE 7-32
Update that violates an integrity constraint
ACCESS USER NOTE Access does not support the CHECK clause. To specify a validation rule in Access, open the table in Design view, and then enter an expression in the column’s Validation Rule property to limit the values that users can enter into the column.
Database Administration
Chapter Summary ●
A view contains data that is derived from existing base tables when users attempt to access the view.
●
To create a view, use the CREATE VIEW command, which includes a defining query that describes the portion of the database included in the view. When a user retrieves data from the view, the DBMS merges the query entered by the user with the defining query and produces the query that the DBMS actually executes.
●
Views provide data independence, allow database access control, and simplify the database structure for users.
●
You cannot update views that involve statistics and views with joins of nonprimary key columns. Updates for these types of views must be made in the base table.
●
Use the DROP VIEW command to delete a view.
●
Use the GRANT command to give users access privileges to data in the database.
●
Use the REVOKE command to terminate previously granted privileges.
●
You can create and use an index to make data retrieval more efficient. Use the CREATE INDEX command to create an index. Use the CREATE UNIQUE INDEX command to enforce a rule so only unique values are allowed in a nonprimary key column.
●
Use the DROP INDEX command to delete an index.
●
The DBMS, not the user, chooses which index to use to accomplish a given task.
●
The DBMS maintains information about the tables, columns, indexes, and other system elements in the system catalog (catalog) or data dictionary. Information about tables is kept in the SYSTABLES table, information about columns is kept in the SYSCOLUMNS table, and information about views is kept in the SYSVIEWS table. In Oracle, these same tables are named DBA_TABLES, DBA_TAB_COLUMNS, and DBA_VIEWS.
●
Use the SELECT command to obtain information from the system catalog. The DBMS updates the system catalog automatically whenever changes are made to the database. In Access, use the Documenter to obtain information about the database objects. SQL Server uses stored procedures to obtain information from the system catalog.
●
Integrity constraints are rules that the data in the database must follow to ensure that only legal values are accepted in specified columns and that primary and foreign key values match between tables. To specify a general integrity constraint, use the CHECK clause. You usually specify primary key constraints when you create a table, but you can specify them later using the ADD PRIMARY KEY clause. To specify a foreign key, use the ADD FOREIGN KEY clause.
226
Key Terms
Chapter 7
ADD FOREIGN KEY
child
ADD PRIMARY KEY
CREATE INDEX
base table
CREATE UNIQUE INDEX
catalog
CREATE VIEW
CHECK
data dictionary
database administration
parent
database administrator
REFERENCES
DBA_TAB_COLUMNS
REVOKE
DBA_TABLES
row-and-column subset view
DBA_VIEWS
security
defining query
SYSCOLUMNS
DROP INDEX
SYSTABLES
DROP VIEW
system catalog
foreign key
SYSVIEWS
GRANT
unique index
index
view
integrity constraint
WITH GRANT OPTION
227
integrity support
Review Questions 1.
What is a view?
2.
Which command creates a view?
3.
What is a defining query?
4.
What happens when a user retrieves data from a view?
5.
What are three advantages of using views?
6.
Which types of views cannot be updated?
7.
Which command deletes a view?
8.
Which command gives users access privileges to various portions of the database?
9.
Which command terminates previously granted privileges?
10.
What is the purpose of an index?
11.
How do you create an index? How do you create a unique index? What is the difference between an index and a unique index?
12.
Which command deletes an index?
13.
Does the DBMS or the user make the choice of which index to use to accomplish a given task?
14.
Describe the information the DBMS maintains in the system catalog. What are the generic names for three tables in the catalog and their corresponding names in Oracle?
15.
Use your favorite Web browser and search engine to find information about a data dictionary. Write a one-page paper that describes other types of information that can be stored in a data dictionary. Be sure to cite the URLs that you use.
16.
Which command do you use to obtain information from the system catalog in Oracle?
17.
How is the system catalog updated?
18.
What are integrity constraints?
19.
How do you specify a general integrity constraint? Database Administration
20.
When would you usually specify primary key constraints? Can you specify them after creating a table? How?
21.
How do you specify a foreign key in Oracle?
22.
Use your favorite Web browser and search engine to find information about referential integrity. Write two or three paragraphs that describe what referential integrity is and include an example of how referential integrity is used in the Premiere Products database. Be sure to cite the URLs that you use.
Exercises 228
Premiere Products Use SQL to make the following changes to the Premiere Products database (see Figure 1-2 in Chapter 1). After each change, execute an appropriate query to show that the change was made correctly. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. For any exercises that use commands not supported by your version of SQL, write the command to accomplish the task. 1.
2.
3.
Chapter 7
Create a view named MAJOR_CUSTOMER. It consists of the customer number, name, balance, credit limit, and rep number for every customer whose credit limit is $10,000 or less. a.
Write and execute the CREATE VIEW command to create the MAJOR_CUSTOMER view.
b.
Write and execute the command to retrieve the customer number and name of each customer in the MAJOR_CUSTOMER view with a balance that exceeds the credit limit.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Create a view named PART_ORDER. It consists of the part number, description, price, order number, order date, number ordered, and quoted price for all order lines currently on file. a.
Write and execute the CREATE VIEW command to create the PART_ORDER view.
b.
Write and execute the command to retrieve the part number, description, order number, and quoted price for all orders in the PART_ORDER view for parts with quoted prices that exceed $100.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Create a view named ORDER_TOTAL. It consists of the order number and order total for each order currently on file. (The order total is the sum of the number of units ordered multiplied by the quoted price on each order line for each order.) Sort the rows by order number. Use TOTAL_AMOUNT as the name for the order total. a.
Write and execute the CREATE VIEW command to create the ORDER_TOTAL view.
b.
Write and execute the command to retrieve the order number and order total for only those orders totaling more than $1,000.
4.
5. 6.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Write, but do not execute, the commands to grant the following privileges: a.
User Ashton must be able to retrieve data from the PART table.
b.
Users Kelly and Morgan must be able to add new orders and order lines.
c.
User James must be able to change the price for all parts.
d.
User Danielson must be able to delete customers.
e.
All users must be able to retrieve each customer’s number, name, street, city, state, and zip code.
f.
User Perez must be able to create an index on the ORDERS table.
g.
User Washington must be able to change the structure of the PART table.
h.
User Grinstead must have all privileges on the ORDERS table.
Write, but do not execute, the command to revoke all privileges from user Ashton. Perform the following tasks: a.
Create an index named PART_INDEX1 on the PART_NUM column in the ORDER_LINE table.
b.
Create an index named PART_INDEX2 on the CLASS column in the PART table.
c.
Create an index named PART_INDEX3 on the CLASS and WAREHOUSE columns in the PART table.
d.
Create an index named PART_INDEX4 on the CLASS and WAREHOUSE columns in the PART table. List item classes in descending order.
7.
Delete the index named PART_INDEX3.
8.
Write the commands to obtain the following information from the system catalog. Do not execute these commands unless your instructor asks you to do so.
9. 10.
229
a.
List every table that contains a column named CUSTOMER_NUM.
b.
List every column in the PART table and its associated data type.
Add the ORDER_NUM column as a foreign key in the ORDER_LINE table. Ensure that the only values entered into the CREDIT_LIMIT column are 5000, 7500, 10000, and 15000.
Henry Books Use SQL to make the following changes to the Henry Books database (Figures 1-4 through 1-7 in Chapter 1). After each change, execute an appropriate query to show that the change was made correctly. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. For any exercises that use commands not supported by your version of SQL, write the command to accomplish the task.
Database Administration
1.
230
2.
3.
4.
5.
Chapter 7
Create a view named PLUME. It consists of the book code, title, type, and price for every book published by the publisher whose code is PL. a.
Write and execute the CREATE VIEW command to create the PLUME view.
b.
Write and execute the command to retrieve the book code, title, and price for every book with a price of less than $13.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Create a view named NONPAPERBACK. It consists of the book code, title, publisher name, and price for every book that is not available in paperback. a.
Write and execute the CREATE VIEW command to create the NONPAPERBACK view.
b.
Write and execute the command to retrieve the book title, publisher name, and price for every book in the NONPAPERBACK view with a price of less than $20.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Create a view named BOOK_INVENTORY. It consists of the branch number and the total number of books on hand for each branch. Use UNITS as the name for the count of books on hand. Group and order the rows by branch number. a.
Write and execute the CREATE VIEW command to create the BOOK_INVENTORY view.
b.
Write and execute the command to retrieve the branch number and units for each branch having more than 25 books on hand.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Write, but do not execute, the commands to grant the following privileges: a.
User Rodriquez must be able to retrieve data from the BOOK table.
b.
Users Gomez and Liston must be able to add new books and publishers to the database.
c.
Users Andrews and Zimmer must be able to change the price of any book.
d.
All users must be able to retrieve the book title, book code, and book price for every book.
e.
User Golden must be able to add and delete publishers.
f.
User Andrews must be able to create an index for the BOOK table.
g.
Users Andrews and Golden must be able to change the structure of the AUTHOR table.
h.
User Golden must have all privileges on the BRANCH, BOOK, and INVENTORY tables.
Write, but do not execute, the command to revoke all privileges from user Andrews.
6.
Create the following indexes: a.
Create an index named BOOK_INDEX1 on the TITLE column in the BOOK table.
b.
Create an index named BOOK_INDEX2 on the TYPE column in the BOOK table.
c.
Create an index named BOOK_INDEX3 on the CITY and PUBLISHER_NAME columns in the PUBLISHER table.
7.
Delete the index named BOOK_INDEX3.
8.
Write the commands to obtain the following information from the system catalog. Do not execute these commands unless your instructor asks you to do so.
9. 10.
231
a.
List every column in the PUBLISHER table and its associated data type.
b.
List every table that contains a column named PUBLISHER_CODE.
c.
List the table name, column name, and data type for the columns named BOOK_CODE, TITLE, and PRICE. Order the results by table name within column name. (That is, column name is the major sort key and table name is the minor sort key.)
Add the PUBLISHER_CODE column as a foreign key in the BOOK table. Ensure that the PAPERBACK column in the BOOK table can accept only values of Y or N.
Alexamara Marina Group Use SQL to make the following changes to the Alexamara Marina Group database (Figures 1-8 through 1-12 in Chapter 1). After each change, execute an appropriate query to show that the change was made correctly. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. For any exercises that use commands not supported by your version of SQL, write the command to accomplish the task. 1.
2.
Create a view named LARGE_SLIP. It consists of the marina number, slip number, rental fee, boat name, and owner number for every slip whose length is 40 feet. a.
Write and execute the CREATE VIEW command to create the LARGE_SLIP view.
b.
Write and execute the command to retrieve the marina number, slip number, rental fee, and boat name for every slip with a rental fee of $3,800 or more.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Create a view named RAY_4025. It consists of the marina number, slip number, length, rental fee, boat name, and owner’s last name for every slip in which the boat type is Ray 4025. a.
Write and execute the CREATE VIEW command to create the RAY_4025 view.
b.
Write and execute the command to retrieve the marina number, slip number, rental fee, boat name, and owner’s last name for every slip in the RAY_4025 view with a rental fee of less than $4,000.
Database Administration
3.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
Create a view named SLIP_FEES. It consists of two columns: the first is the slip length, and the second is the average fee for all slips in the MARINA_SLIP table that have that length. Use AVERAGE_FEE as the name for the average fee. Group and order the rows by slip length. a.
Write and execute the CREATE VIEW command to create the SLIP_FEES view.
b.
Write and execute the command to retrieve the slip length and average fee for each length for which the average fee is less than $3,500.
c.
Write and execute the query that the DBMS actually executes.
d.
Does updating the database through this view create any problems? If so, what are they? If not, why not?
232
4.
a.
User Oliver must be able to retrieve data from the MARINA_SLIP table.
b.
Users Crandall and Perez must be able to add new owners and slips to the database.
c.
Users Johnson and Klein must be able to change the rental fee of any slip.
d.
All users must be able to retrieve the length, boat name, and owner number for every slip.
e.
User Klein must be able to add and delete service categories.
f.
User Adams must be able to create an index on the SERVICE_REQUEST table.
g.
Users Adams and Klein must be able to change the structure of the MARINA_SLIP table.
h.
User Klein must have all privileges on the MARINA, OWNER, and MARINA_SLIP tables.
5.
Write, but do not execute, the command to revoke all privileges from user Adams.
6.
Create the following indexes: a.
Create an index named BOAT_INDEX1 on the OWNER_NUM column in the MARINA_SLIP table.
b.
Create an index named BOAT_INDEX2 on the BOAT_NAME column in the MARINA_SLIP table.
c.
Create an index named BOAT_INDEX3 on the LENGTH and BOAT_NAME columns in the MARINA_SLIP table. List the lengths in descending order.
7.
Delete the index named BOAT_INDEX3; it is no longer necessary.
8.
Write the commands to obtain the following information from the system catalog. Do not execute these commands unless your instructor specifically asks you to do so.
9. 10.
Chapter 7
Write, but do not execute, the commands to grant the following privileges:
a.
List every column in the MARINA_SLIP table and its associated data type.
b.
List every table and view that contains a column named MARINA_NUM.
Add the OWNER_NUM column as a foreign key in the MARINA_SLIP table. Ensure that the LENGTH column in the MARINA_SLIP table can accept only values of 25, 30, or 40.
CHAPTER
8
S Q L F UN CTION S AN D P ROCED URES LEARNING
OBJECTIVES
Objectives ●
Understand how to use functions in queries
●
Use the UPPER and LOWER functions with character data
●
Use the ROUND and FLOOR functions with numeric data
●
Add a specific number of months or days to a date
●
Calculate the number of days between two dates
●
Use concatenation in a query
●
Embed SQL commands in PL/SQL and T-SQL procedures
●
Retrieve single rows using embedded SQL
●
Update a table using embedded INSERT, UPDATE, and DELETE commands
●
Use cursors to retrieve multiple rows in embedded SQL
●
Manage errors in procedures containing embedded SQL commands
●
Use SQL in a language that does not support embedded SQL commands
●
Use triggers
INTRODUCTION You already have used functions that apply to groups (such as SUM and AVG). In this chapter, you will learn to use functions that apply to values in individual rows. Specifically, you will see how to use functions with characters or text, numbers, and dates. You will learn how to concatenate values in a query. You will embed SQL commands in PL/SQL and T-SQL procedures to retrieve rows and update data. You will
examine the different ways to manage errors in procedures. Finally, you will learn how to create and use cursors and triggers.
USING SQL IN A PROGRAMMING ENVIRONMENT
234
SQL is a powerful nonprocedural language in which you communicate tasks to the computer using simple commands. As in other nonprocedural languages, you can accomplish many tasks using a single command. Although SQL and other nonprocedural languages are wellequipped to store and query data, sometimes you might need to complete tasks that are beyond the capabilities of SQL. In such cases, you need to use a procedural language. A procedural language is one in which you must give the computer the step-by-step process for accomplishing a task. PL/SQL, which was developed by Oracle as an extension of SQL, is an example of a procedural language. This chapter uses PL/SQL to illustrate how to use SQL in a programming environment by embedding SQL commands in another language. The examples in this chapter illustrate how to use embedded SQL commands to retrieve a single row, insert new rows, update and delete existing rows, and retrieve multiple rows. In the process, you will create stored procedures that are saved and are available for use at any time. T-SQL, which stands for Transact-SQL, is another extension of SQL. T-SQL is the procedural language that SQL Server uses. You can perform tasks, such as retrieving a single row, inserting new rows, and retrieving multiple rows, using T-SQL in SQL Server. Although the language syntax is slightly different in T-SQL when compared to PL/SQL, the functionality and the results are the same. You cannot embed SQL commands in Access programs the way you can in PL/SQL and T-SQL. There are ways to use the commands, however, as you’ll learn later in this chapter.
NOTE This chapter assumes that you have some programming background and does not cover programming basics. To understand the first part of this chapter, you should be familiar with variables, declaring variables, and creating procedural code, including IF statements and loops. To understand the Access section at the end of the chapter, you should be familiar with Function and Sub procedures, and the process for sequentially accessing all records in a recordset, such as using a loop to process all the records in a table.
ACCESS USER NOTE If you are using Access, you will not be able to complete the material in this chapter that deals with PL/SQL and T-SQL procedures. Be sure to read this information so you will understand these important concepts. You will, however, be able to complete the steps in the “Using SQL in Microsoft Access” section.
Chapter 8
USING FUNCTIONS You already have used aggregate functions to perform calculations based on groups of records. For example, SUM(BALANCE) calculates the sum of the balances on all records that satisfy the condition in the WHERE clause. When you use a GROUP BY clause, the DBMS will calculate the sum for each record in a group. SQL also includes functions that affect single records. Some functions affect character data and others let you manipulate numeric data. The supported SQL functions vary between SQL implementations. This section will illustrate some common functions. For additional information about the functions your SQL implementation supports, consult the program’s documentation. 235
Character Functions SQL includes several functions that affect character data. Example 1 illustrates the use of the UPPER function.
EXAMPLE 1 List the rep number and last name for each sales rep. Display the last name in uppercase letters.
The UPPER function displays a value in uppercase letters; for example, the function UPPER(LAST_NAME) displays the last name Kaiser as KAISER. (Note that the UPPER function simply displays the last name in uppercase letters; it does not change the last name stored in the table to uppercase letters.) The item in parentheses (LAST_NAME) is called the argument for the function. The value produced by the function is the result of displaying all lowercase letters in the value stored in the LAST_NAME column in uppercase letters. The query and its results are shown in Figure 8-1.
Argument
UPPER function
Last names displayed in uppercase
FIGURE 8-1
Using the UPPER function to display character data in uppercase letters
You can use functions in WHERE clauses as well. For example, the condition UPPER(LAST_NAME) = 'KAISER' would be true for names like Kaiser, KAISER, and KaIsER, SQL Functions and Procedures
because the result of applying the UPPER function to any of these values would result in the value KAISER. To display a value in lowercase letters, you can use the LOWER function. SQL Server supports both the UPPER and LOWER function.
ACCESS USER NOTE In Access, the UCASE() function displays a value in uppercase letters and the LCASE() function displays a value in lowercase letters. For example, if the value stored in the LAST_NAME column is Kaiser, UCASE(LAST_NAME) would result in the value KAISER and LCASE(LAST_NAME) would result in the value kaiser. 236
Number Functions SQL also includes functions that affect numeric data. The ROUND function, which rounds values to a specified number of decimal places, is illustrated in Example 2.
EXAMPLE 2 List the part number and price for all parts. Round the price to the nearest whole dollar amount.
A function can have more than one argument. The ROUND function, which rounds a numeric value to a desired number of decimal places, has two arguments. The first argument is the value to be rounded; the second argument indicates the number of decimal places to which to round the result. For example, ROUND(PRICE,0) will round the values in the PRICE column to zero decimal places (a whole number). If a price is 24.95, the result will be 25. If the price is 24.25, on the other hand, the result will be 24. Figure 8-2 shows the query and results to round values in the PRICE column to zero decimal places. The computed column ROUND(PRICE,0) is named ROUNDED_PRICE.
Chapter 8
ROUND function
Price values rounded to zero decimal places in a column named ROUNDED_PRICE
FIGURE 8-2
237
Using the ROUND function to round numeric values
Rather than rounding (using the ROUND function), you might need to truncate (remove) everything to the right of the decimal point. To do so, use the FLOOR function, which has only one argument. If a price is 24.95, for example, ROUND(PRICE,0) would result in 25, whereas FLOOR(PRICE) would result in 24. SQL Server supports both the ROUND and the FLOOR functions. Microsoft Access supports only the ROUND function.
Working with Dates SQL uses functions and calculations for manipulating dates. To add a specific number of months to a date, you can use the ADD_MONTHS function as illustrated in Example 3.
EXAMPLE 3 For each order, list the order number and the date that is two months after the order date. Name this date FUTURE_DATE.
The ADD_MONTHS function has two arguments. The first argument is the date to which you want to add a specific number of months, and the second argument is the number of months. To add two months to the order date, for example, the expression is ADD_MONTHS(ORDER_DATE,2) as illustrated in Figure 8-3.
SQL Functions and Procedures
ADD_MONTHS function
Two months added to each order date in a column named FUTURE_DATE
238
FIGURE 8-3
Using the ADD_MONTHS function to add months to a date
ACCESS USER NOTE To add a number of months to a date in Access, use the DATEADD() function, which has three arguments. The first argument includes the interval of time to be added; the letter “m” indicates that months will be added. The second argument includes the number of intervals to be added. The third argument includes the date to be manipulated. For example, to add two months to the dates stored in the ORDER_DATE column, the appropriate function would be DATEADD('m', 2, ORDER_DATE).
SQL SERVER USER NOTE To add a number of months to a date in SQL Server, use the DATEADD() function, which has three arguments. The first argument includes the interval of time to be added; the letter “m” indicates that months will be added. The second argument includes the number of intervals to be added. The third argument includes the date to be manipulated. For example, to add two months to the dates stored in the ORDER_DATE column, the appropriate function would be DATEADD('m', 2, ORDER_DATE).
EXAMPLE 4 For each order, list the order number and the date that is seven days after the order date. Name this date FUTURE_DATE.
To add a specific number of days to a date, you do not need a function. You can add the number of days to the order date as illustrated in Figure 8-4. (You can also subtract dates in the same way.) This method works in Oracle, Access, and SQL Server.
Chapter 8
Expression that adds seven days to the order dates
Seven days added to each order date
239
FIGURE 8-4
Adding days to dates
EXAMPLE 5 For each order, list the order number, today’s date, the order date, and the number of days between the order date and today’s date. Name today’s date TODAYS_DATE and name the number of days between the order date and today’s date DAYS_PAST.
You can use the SYSDATE function to obtain today’s date, as shown in Figure 8-5. The command in the figure uses SYSDATE to display today’s date and also uses SYSDATE in a computation to determine the number of days between the order date and today’s date. The values for DAYS_PAST include decimal places. You could remove these decimal places by using the ROUND or FLOOR functions, if desired.
SQL Functions and Procedures
Number of days between today's date and the order date Expression to calculate the number of days between today's date and the order date
240
FIGURE 8-5
Calculating the number of days between two dates
ACCESS USER NOTE In Access, use the DATE() function to obtain today’s date, rather than SYSDATE. The DATE() function has no arguments, so you would write DATE() in place of SYSDATE.
SQL SERVER USER NOTE In SQL Server, use the GETDATE() function to obtain today’s date, rather than SYSDATE. The GETDATE() function has no arguments, so you would write GETDATE() in place of SYSDATE.
CONCATENATING COLUMNS Sometimes you need to concatenate, or combine, two or more character columns into a single expression when displaying them in a query; the process is called concatenation. To concatenate columns, you type two vertical lines (||) between the column names, as illustrated in Example 6.
EXAMPLE 6 List the number and name of each sales rep. Concatenate the FIRST_NAME and LAST_NAME columns into a single value, with a space separating the first and last names.
To concatenate the FIRST_NAME and LAST_NAME columns, the expression is FIRST_NAME||LAST_NAME. When the first name doesn’t include sufficient characters to fill the width of the column (as determined by the number of characters specified in the CREATE TABLE command), SQL inserts extra spaces. For example, when the Chapter 8
FIRST_NAME column is 12 characters wide, the first name is Mary, and the last name is Johnson, the expression FIRST_NAME||LAST_NAME appears as Mary, followed by eight spaces, and then Johnson. To remove the extra spaces following the first name value, you use the RTRIM (right trim) function. When you apply this function to the value in a column, SQL displays the original value and removes any spaces inserted at the end of the value. Figure 8-6 shows the query and output with the extra spaces removed. For sales rep 20, for example, this command trims the first name to “Valerie,” concatenates it with a single space, and then concatenates the last name “Kaiser.”
RTRIM function
Concatenation symbol
241
Concatenated names
FIGURE 8-6
Concatenating two columns and using the RTRIM function
Q&A Question: Why is it necessary to insert a single space character in single quotation marks in the query? Answer: Without the space character, there would be no space between the first and last names. The name of sales rep 20, for example, would be displayed as “ValerieKaiser.”
ACCESS USER NOTE In Access, use the & symbol to concatenate columns. It is not necessary to trim the columns because Access will trim them automatically. The corresponding query in Access is: SELECT REP_NUM, FIRST_NAME&' '&LAST_NAME FROM REP;
SQL Functions and Procedures
SQL SERVER USER NOTE In SQL Server, use the + symbol to concatenate columns. The corresponding query in SQL Server is: SELECT REP_NUM, RTRIM(FIRST_NAME)+' '+RTRIM(LAST_NAME) FROM REP;
STORED PROCEDURES
242
In a client/server system, the database is stored on a computer called the server and users access the database through clients. A client is a computer that is connected to a network and has access through the server to the database. Every time a user executes a query, the DBMS must determine the best way to process the query and provide the results. For example, the DBMS must determine which indexes are available and whether it can use those indexes to make the processing of the query more efficient. When you anticipate running a particular query often, you can improve overall performance by saving the query in a file called a stored procedure. The stored procedure is placed on the server. The DBMS compiles the stored procedure (translating it into machine code) and creates an execution plan, which is the most efficient way of obtaining the results. From that point on, users execute the compiled, optimized code in the stored procedure. Another reason for saving a query as a stored procedure, even when you are not working in a client/server system, is convenience. Rather than retyping the entire query each time you need it, you can use the stored procedure. For example, suppose you frequently execute a query that selects a sales rep with a given number and then displays the concatenation of the first name and last name of the sales rep. Instead of typing the query each time you want to display a sales rep’s name, you can store the query in a stored procedure. You would then only need to run the stored procedure when you want to display a sales rep’s name.
ACCESS USER NOTE Although Access does not support stored procedures, you can achieve some of the same convenience by creating a parameter query that prompts the user for the arguments you would otherwise use in a stored procedure.
In Oracle, you create stored procedures using a language called PL/SQL. You create and save the procedures as script files.
Retrieving a Single Row and Column Example 7 illustrates using a stored procedure to retrieve a single row and column from a table.
Chapter 8
EXAMPLE 7 Write a PL/SQL procedure that takes a rep number as input and displays the corresponding rep name.
Figure 8-7 shows a procedure to find the name of the representative whose number is stored in the I_REP_NUM argument. Because the restriction involves the primary key, the query will produce only one row of output. (You will see how to handle queries whose results can contain multiple rows later in this chapter.) The command shown in Figure 8-7 is stored in a script file and is displayed in the Script Editor. To create the procedure, you would run the script file. Assuming that the script file does not contain any errors, Oracle would then create the procedure and it would be available for use.
Local variables
Procedure name
243
Argument
Command to select LAST_NAME and FIRST_NAME and place them in local variables
Expression to output the rep’s name
FIGURE 8-7
Procedure to find a rep’s name given the rep’s number
NOTE PL/SQL commands, like SQL commands, are free-format and can include blank lines to separate important sections of the procedure and spaces on the lines to make the commands more readable.
The CREATE PROCEDURE command in the stored procedure causes Oracle to create a procedure named DISP_REP_NAME. By including the optional OR REPLACE clause in the CREATE PROCEDURE command, you can use the command to modify an existing procedure. If you omit the OR REPLACE clause, you would need to drop the procedure and then re-create it in order to change the procedure later. The first line of the command contains a single argument, I_REP_NUM. The word IN following the single argument name indicates that I_REP_NUM will be used for input. That SQL Functions and Procedures
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is, the user must enter a value for I_REP_NUM to use the procedure. Other possibilities are OUT, which indicates that the procedure will set a value for the argument, and INOUT, which indicates that the user will enter a value that the procedure can later change. Variable names in PL/SQL must start with a letter and can contain letters, dollar signs, underscores, and number signs, but cannot exceed 30 characters. When declaring variables, you must assign the variable a data type, just as you do in the SQL CREATE TABLE command. You can ensure that a variable has the same data type as a particular column in a table by using the %TYPE attribute. To do so, you include the name of the table, followed by a period and the name of the column, and then %TYPE. When you use %TYPE, you do not enter a data type because the variable is automatically assigned the same type as the corresponding column. In the first line of the script file shown in Figure 8-7, assigning the variable I_REP_NUM the same type as the REP_NUM column in the REP table is written as REP.REP_NUM%TYPE. The first line of the CREATE PROCEDURE command ends with the word AS and is followed by the commands in the procedure. The commands on lines 2 and 3 declare the local variables the procedure requires. In Figure 8-7, lines 2 and 3 create two variables named I_LAST_NAME and I_FIRST_NAME. Both variables are assigned data types using %TYPE. The procedural code, which contains the commands that specify the procedure’s function, appears between the BEGIN and END commands. In Figure 8-7, the procedural code begins with the SQL command to select the last name and first name of the sales rep whose number is stored in I_REP_NUM. The SQL command uses the INTO clause to place the results in the I_LAST_NAME and I_FIRST_NAME variables. The next command uses the DBMS_OUTPUT.PUT_LINE procedure to display the concatenation of the trimmed I_FIRST_NAME and I_LAST_NAME variables. Notice that a semicolon ends each variable declaration, command, and the word END. The slash (/) at the end of the procedure appears on its own line. In some Oracle environments, the slash is optional. A good practice is to include the slash, even when it’s not necessary, so your procedure will always work correctly.
NOTE DBMS_OUTPUT is a package that contains multiple procedures, including PUT_LINE. The SQL Commands page automatically displays the output produced by DBMS_OUTPUT.
To call (or use) the procedure from the SQL Commands page, type the word BEGIN, followed by the name of the procedure including the desired value for the argument in parentheses, followed by the word END, a semicolon, and a slash on a separate line. To use the DISP_REP_NAME procedure to find the name of sales rep 20, for example, type the command shown in Figure 8-8.
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Value for argument (sales rep number 20)
Command to use the DISP_REP_NAME procedure
Procedure result (name of sales rep 20)
FIGURE 8-8
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Using the DISP_REP_NAME procedure within an SQL command
ERROR HANDLING Procedures must be able to handle conditions that can arise when accessing the database. For example, the user enters a rep number and the DISP_REP_NAME procedure displays the corresponding rep’s name. What happens when the user enters an invalid rep number? This situation results in the error message shown in Figure 8-9 because Oracle will not find any last name to display.
Invalid rep number
System error message
FIGURE 8-9
System error that occurs when a user enters an invalid rep number
You can include the EXCEPTION clause shown in Figure 8-10 to handle processing an invalid rep number. When a user enters a rep number that does not match any rep number in the REP table, the NO_DATA_FOUND condition on line 13 will be true. When the NO_DATA_FOUND condition is true, the procedure displays the “No rep with this number:” message followed by the invalid rep number.
SQL Functions and Procedures
EXCEPTION clause
Tests for NO_DATA_FOUND condition
Action to take when no data is found
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FIGURE 8-10
PL/SQL procedure with error handling
When you use this version of the procedure and enter an invalid rep number, you will see the error message from the procedure (Figure 8-11) instead of the system error message (Figure 8-9).
Invalid rep number
Error message from the procedure
FIGURE 8-11
Error message that occurs when a user enters an invalid rep number
The DISP_REP_NAME procedure must handle an error that results when a user enters an invalid rep number. There are other types of errors that procedures must handle, depending on the processing required. For example, a user might enter a commission rate in a procedure to find the name of the sales rep who has that commission rate. When the user enters the rate 0.05, the procedure will display the TOO_MANY_ROWS error because Valerie Kaiser and Juan Perez both have this same commission rate—the procedure finds two rows instead of one. You can manage this error by writing a WHEN clause that contains a TOO_MANY_ROWS condition, following the EXCEPTION clause in the procedure. You can write both WHEN clauses in the same procedure or in separate procedures. When adding both WHEN clauses to the same procedure, however, the EXCEPTION clause appears only once.
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USING UPDATE PROCEDURES In Chapter 6, you learned how to use SQL commands to update data. You can use the same commands within procedures. A procedure that updates data is called an update procedure.
Changing Data with a Procedure You can use an update procedure to change a row in a table, as illustrated in Example 8.
EXAMPLE 8 Change the name of the customer whose number is stored in I_CUSTOMER_NUM to the value currently stored in I_CUSTOMER_NAME.
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This procedure is similar to the procedures used in previous examples with two main differences: it uses an UPDATE command instead of a SELECT command, and there are two arguments, I_CUSTOMER_NUM and I_CUSTOMER_NAME. The I_CUSTOMER_NUM argument stores the customer number to be updated and the I_CUSTOMER_NAME argument stores the new value for the customer name. The procedure appears in Figure 8-12.
Arguments
FIGURE 8-12
Using a procedure to update a row
When you run this procedure, you will need to furnish values for two arguments. Figure 8-13 uses this procedure to change the name of customer 725 to Deerfield’s.
Argument that stores the new name
Argument that stores the customer number
FIGURE 8-13
Using a procedure to update the name of customer 725
SQL Functions and Procedures
Deleting Data with a Procedure Just as you would expect, if you can use an update procedure to change a rows in a table, you can also use one to delete a row from a table, as illustrated in Example 9.
EXAMPLE 9 Delete the order whose number is stored in I_ORDER_NUM from the ORDERS table, and also delete each order line for the order whose order number is currently stored in the variable from the ORDER_LINE table. 248
If you attempt to delete the order in the ORDERS table first, referential integrity will prevent the deletion because matching rows would still exist in the ORDER_LINE table, so it is a good idea to delete the orders from the ORDER_LINE table first. The procedure to delete an order and its related order lines appears in Figure 8-14. This procedure contains two DELETE commands. The first command deletes all order lines in the ORDER_LINE table on which the order number matches the value stored in the I_ORDER_NUM argument. The second command deletes the order in the ORDERS table whose order number matches the value stored in the I_ORDER_NUM argument.
Argument that stores the order number to delete
Command to delete all rows in the ORDER_LINE table that matches the entered order number Command to delete the row in the ORDERS table that matches the entered order number
FIGURE 8-14
Procedure to delete a row and related rows from multiple tables
Figure 8-15 shows the use of this procedure to delete order number 21610. Even though there are two DELETE commands in the procedure, the user enters the order number only once.
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FIGURE 8-15
Using the procedure to delete an order 249
SELECTING MULTIPLE ROWS WITH A PROCEDURE The procedures you have seen so far include commands that retrieve individual rows. You can use an UPDATE or a DELETE command in PL/SQL to update or delete multiple rows. The commands are executed and the updates or deletions occur. Then the procedure can move on to the next task. What happens when a SELECT command in a procedure retrieves multiple rows? For example, suppose the SELECT command retrieves the number and name of each customer represented by the sales rep whose number is stored in I_REP_NUM. There is a problem—PL/SQL can process only one record at a time, but this SQL command retrieves more than one row. Whose number and name is placed in I_CUSTOMER_NUM and I_CUSTOMER_NAME when the command retrieves more than one customer row? Should you make I_CUSTOMER_NUM and I_CUSTOMER_NAME arrays capable of holding multiple rows and, if so, what should be the size of these arrays? Fortunately, you can solve this problem by using a cursor.
Using a Cursor A cursor is a pointer to a row in the collection of rows retrieved by an SQL command. (This is not the same cursor that you see on your computer screen.) The cursor advances one row at a time to provide sequential, one-record-at-a-time access to the retrieved rows so PL/SQL can process the rows. By using a cursor, PL/SQL can process the set of retrieved rows as though they were records in a sequential file. To use a cursor, you must first declare it, as illustrated in Example 10.
EXAMPLE 10 Retrieve and list the number and name of each customer represented by the sales rep whose number is stored in the variable I_REP_NUM.
SQL Functions and Procedures
The first step in using a cursor is to declare the cursor and describe the associated query in the declaration section of the procedure. In this example, assuming the cursor is named CUSTGROUP, the command to declare the cursor is: CURSOR CUSTGROUP IS SELECT CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER WHERE REP_NUM = I_REP_NUM;
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This command does not cause the query to be executed at this time; it only declares a cursor named CUSTGROUP and associates the cursor with the indicated query. Using a cursor in a procedure involves three commands: OPEN, FETCH, and CLOSE. The OPEN command opens the cursor and causes the query to be executed, making the results available to the procedure. Executing a FETCH command advances the cursor to the next row in the set of rows retrieved by the query and places the contents of the row in the indicated variables. Finally, the CLOSE command closes a cursor and deactivates it. Data retrieved by the execution of the query is no longer available. The cursor could be opened again later and processing could begin again. The OPEN, FETCH, and CLOSE commands used in processing a cursor are analogous to the OPEN, READ, and CLOSE commands used in processing a sequential file.
Opening a Cursor Prior to opening the cursor, there are no rows available to be fetched. In Figure 8-16, this is indicated by the absence of data in the CUSTGROUP portion of the figure. The right side of the figure illustrates the variables into which the data will be placed (I_CUSTOMER_NUM and I_CUSTOMER_NAME) and the value CUSTGROUP%NOTFOUND. Once the cursor has been opened and all the records have been fetched, the CUSTGROUP%NOTFOUND value is set to TRUE. Procedures using the cursor can use this value to indicate when the fetching of rows is complete. CUSTGROUP CUSTOMER_ NUM
I_CUSTOMER_ NUM
CUSTOMER_NAME no row to be fetched
FIGURE 8-16
I_CUSTOMER_NAME
CUSTGROUP %NOTFOUND FALSE
Before OPEN
The OPEN command is written as follows: OPEN CUSTGROUP;
Figure 8-17 shows the result of opening the CUSTGROUP cursor. In the figure, assume that I_REP_NUM is set to 20 before the OPEN command is executed; there are now three rows available to be fetched. No rows have yet been fetched, as indicated by the absence of values in I_CUSTOMER_NUM and I_CUSTOMER_NAME. CUSTGROUP%NOTFOUND is still FALSE. The cursor is positioned at the first row; that is, the next FETCH command causes the contents of the first row to be placed in the indicated variables. Chapter 8
CUSTGROUP CUSTOMER_ NUM 148
Al's Appliance and Sport
524
Kline's
842
All Season
FIGURE 8-17
I_CUSTOMER_ NUM
CUSTOMER_NAME
I_CUSTOMER_NAME
next row to be fetched
CUSTGROUP %NOTFOUND FALSE
After OPEN, but before first FETCH
Fetching Rows from a Cursor To fetch (get) the next row from a cursor, use the FETCH command. The FETCH command is written as follows:
251
FETCH CUSTGROUP INTO I_CUSTOMER_NUM, I_CUSTOMER_NAME;
Note that the INTO clause is associated with the FETCH command itself and not with the query used in the cursor definition. The execution of this query could produce multiple rows. The execution of the FETCH command produces only a single row, so it is appropriate that the FETCH command causes data to be placed in the indicated variables. Figure 8-18 through Figure 8-21 show the result of four FETCH commands. The first three fetches are successful. In each case, the data from the appropriate row in the cursor is placed in the indicated variables and CUSTGROUP%NOTFOUND is still FALSE. The fourth FETCH command is different, however, because there is no more data to fetch. In this case, the contents of the variables are left untouched and CUSTGROUP%NOTFOUND is set to TRUE. CUSTGROUP CUSTOMER_ NUM 148
CUSTOMER_NAME Al's Appliance and Sport
524
Kline's
842
All Season
FIGURE 8-18
I_CUSTOMER_ NUM 148
I_CUSTOMER_NAME
I_CUSTOMER_ NUM 524
I_CUSTOMER_NAME
Al's Appliance and Sport
CUSTGROUP %NOTFOUND FALSE
next row to be fetched
After first FETCH
CUSTGROUP CUSTOMER_ NUM 148
CUSTOMER_NAME Al's Appliance and Sport
524
Kline's
842
All Season
FIGURE 8-19
Kline's
CUSTGROUP %NOTFOUND FALSE
next row to be fetched
After second FETCH
SQL Functions and Procedures
CUSTGROUP CUSTOMER_ NUM 148
I_CUSTOMER_ NUM 842
CUSTOMER_NAME Al's Appliance and Sport
524
Kline's
842
All Season
I_CUSTOMER_NAME All Season
CUSTGROUP %NOTFOUND FALSE
next row to be fetched
FIGURE 8-20
After third FETCH
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CUSTGROUP CUSTOMER_ NUM 148
I_CUSTOMER_ NUM 842
CUSTOMER_NAME Al's Appliance and Sport
524
Kline's
842
All Season
I_CUSTOMER_NAME All Season
CUSTGROUP %NOTFOUND TRUE
no more rows to be fetched
FIGURE 8-21
After attempting a fourth FETCH (CUSTGROUP%NOTFOUND is TRUE)
Closing a Cursor The CLOSE command is written as follows: CLOSE CUSTGROUP;
Figure 8-22 shows the result of closing the CUSTGROUP cursor. The data is no longer available. CUSTGROUP CUSTOMER_ NUM
CUSTOMER_NAME no rows to be fetched
FIGURE 8-22
After CLOSE
Writing a Complete Procedure Using a Cursor Figure 8-23 shows a complete procedure using a cursor. The declaration portion contains the CUSTGROUP cursor definition. The procedural portion begins with the command to open the CUSTGROUP cursor. The statements between the LOOP and END LOOP commands create a loop that begins by fetching the next row from the cursor and placing the
Chapter 8
results in I_CUSTOMER_NUM and I_CUSTOMER_NAME. The EXIT command tests the condition CUSTGROUP%NOTFOUND. If the condition is true, the loop is terminated. If the condition is not true, the DBMS_OUTPUT.PUT_LINE commands display the contents of I_CUSTOMER_NUM and I_CUSTOMER_NAME.
Cursor declaration Command to fetch a row from the cursor
Command to display the results
FIGURE 8-23
253
Command to determine whether a row was successfully fetched
Procedure with a cursor
Figure 8-24 shows the results of using the procedure. After the user enters 20 as the value for the rep number, the procedure displays the number and name of each customer of sales rep 20.
20 entered as the argument for the rep number
Customers of rep 20
FIGURE 8-24
Results of using the procedure
SQL Functions and Procedures
Using More Complex Cursors The query formulation that defined the cursor in Example 10 was straightforward. Any SQL query is legitimate in a cursor definition. In fact, the more complicated the requirements for retrieval, the more numerous the benefits derived by the programmer who uses embedded SQL. Consider the query in Example 11.
EXAMPLE 11
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For each order that contains an order line for the part whose part number is stored in I_PART_NUM, retrieve the order number, order date, customer number, name of the customer that placed the order, and last and first names of the sales rep who represents the customer.
Opening and closing the cursor is done exactly as shown in Example 10. The only difference in the FETCH command is that a different set of variables is used in the INTO clause. Thus, the only real difference is the cursor definition. The procedure shown in Figure 8-25 contains the appropriate cursor definition.
FIGURE 8-25
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Procedure with a cursor that involves joining multiple tables
The results of using this procedure to display the results for part DR93 are shown in Figure 8-26.
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FIGURE 8-26
Results of using the procedure to display orders containing part DR93
Advantages of Cursors The retrieval requirements in Example 11 are substantial. Beyond coding the preceding cursor definition, the programmer doesn’t need to worry about the mechanics of obtaining the necessary data or placing it in the right order, because this happens automatically when the cursor is opened. To the programmer, it seems as if a sequential file already exists that contains the correct data, sorted in the right order. This assumption leads to three main advantages: 1. 2.
3.
The coding in the procedure is greatly simplified. In a normal program, the programmer must determine the most efficient way to access the data. In a program or procedure using embedded SQL, the optimizer determines the best way to access the data. The programmer isn’t concerned with the best way to retrieve the data. In addition, when an underlying structure changes (for example, an additional index is created), the optimizer determines the best way to execute the query with the new structure. The program or procedure does not have to change at all. When the database structure changes in such a way that the necessary information is still obtainable using a different query, the only change required in the program or procedure is the cursor definition. The procedural code is not affected.
SQL Functions and Procedures
U S I N G T- S Q L I N S Q L S E R V E R SQL Server uses an extended version of SQL called T-SQL (Transact-SQL). You can use T-SQL to create stored procedures and use cursors. The reasons for creating and using stored procedures and cursors are identical to those discussed in the PL/SQL section. Only the command syntax is different.
Retrieving a Single Row and Column In Example 7, you learned how to write a procedure in PL/SQL that takes a rep number as input and displays the corresponding rep name. The following code shows how you would create the stored procedure in T-SQL: 256
CREATE PROCEDURE usp_DISP_REP_NAME @repnum char(2) AS SELECT RTRIM(FIRST_NAME)+' '+RTRIM(LAST_NAME) FROM REP WHERE REP_NUM = @repnum
The CREATE PROCEDURE command in the stored procedure causes SQL Server to create a procedure named usp_DISP_REP_NAME. The usp_ prefix identifies the procedure as a user-stored procedure. Although using the prefix is optional, it is an easy way to differentiate user-stored procedures from SQL Server system-stored procedures. The argument for this procedure is @repnum. In T-SQL, you must assign a data type to parameters. All arguments start with the at (@) sign. Arguments should have the same data type and length as the particular column in a table that they represent. In the REP table, REP_NUM was defined with a CHAR data type and a length of 2. The CREATE PROCEDURE ends with the word AS followed by the SELECT command that comprises the procedure. To call the procedure, use the EXEC command and include any arguments in single quotes. The procedure to find the name of sales rep 20 is: EXEC usp_DISP_REP_NAME'20'
The result of executing this procedure is the same as that shown in Figure 8-8.
Changing Data with a Stored Procedure In Example 8, you learned how to write a procedure in PL/SQL that changes the name of a customer. The following commands show how to create the stored procedure in T-SQL: CREATE PROCEDURE usp_CHG_CUST_NAME @custnum char(3), @custname char(35) AS UPDATE CUSTOMER SET CUSTOMER_NAME = @custname WHERE CUSTOMER_NUM = @custnum
The procedure has two arguments, @custnum and @custname, and uses an UPDATE command instead of a SELECT command. To execute a stored procedure with two arguments, separate the arguments with a comma as shown in the following command: EXEC usp_CHG_CUST_NAME'725','Deerfield's' Chapter 8
Deleting Data with a Stored Procedure In Example 9, you learned how to write a procedure in PL/SQL that deletes an order number from both the ORDER_LINE table and the ORDERS table. The following commands show how to create the stored procedure in T-SQL: CREATE PROCEDURE usp_DEL_ORDER @ordernum char(5) AS DELETE FROM ORDER_LINE WHERE ORDER_NUM = @ordernum 257
DELETE FROM ORDERS WHERE ORDER_NUM = @ordernum
Using a Cursor Cursors serve the same purpose in T-SQL as they do in PL/SQL and work exactly the same way. You need to declare a cursor, open a cursor, fetch rows from a cursor, and close a cursor. The only difference is in the command syntax. The following T-SQL code performs exactly the same task as that shown in Example 10: CREATE PROCEDURE usp_DISP_REP_CUST @repnum char(2) AS DECLARE @custnum char(3) DECLARE @custname char(35) DECLARE mycursor CURSOR READ_ONLY FOR SELECT CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER WHERE REP_NUM = @repnum OPEN mycursor FETCH NEXT FROM mycursor INTO @custnum, @custname WHILE @@FETCH_STATUS = 0 BEGIN PRINT @custnum+' '[email protected] FETCH NEXT FROM mycursor INTO @custnum, @custname END CLOSE mycursor DEALLOCATE mycursor
The procedure uses one argument, @repnum. It also uses two variables, and each variable must be declared using a DECLARE statement. You also declare the cursor by giving it a
SQL Functions and Procedures
name, describing its properties, and associating it with a SELECT statement. The cursor property, READ_ONLY, means that the cursor is used for retrieval purposes only. The OPEN, FETCH, and CLOSE commands perform exactly the same tasks in T-SQL as they do in PL/SQL. The OPEN command opens the cursor and causes the query to be executed. The FETCH command advances the cursor to the next row and places the contents of the row in the indicated variables. The CLOSE command closes a cursor and the DEALLOCATE command deletes the cursor. The DEALLOCATE command is not necessary but it does enable the user to use the same cursor name with another procedure. The WHILE loop will repeat until the value of the system variable @@FETCH_STATUS is not zero. The PRINT command will output the values stored in the @custnum and @custname variables. 258
Using More Complex Cursors T-SQL also can handle more complex queries. The T-SQL code for Example 11 is shown below: CREATE PROCEDURE usp_DISP_PART_ORDERS @partnum char(4) AS DECLARE @ordernum char(5) DECLARE @orderdate datetime DECLARE @custnum char(3) DECLARE @repnum char(2) DECLARE @lastname char(15) DECLARE @firstname char(15) DECLARE mycursor CURSOR READ_ONLY FOR SELECT ORDERS.ORDER_NUM, ORDER_DATE, ORDERS.CUSTOMER_NUM, CUSTOMER.REP_NUM, LAST_NAME, FIRST_NAME FROM ORDER_LINE, ORDERS, CUSTOMER, REP WHERE ORDER_LINE.ORDER_NUM = ORDERS.ORDER_NUM AND ORDERS.CUSTOMER_NUM = CUSTOMER.CUSTOMER_NUM AND CUSTOMER.REP_NUM = REP.REP_NUM AND PART_NUM = @partnum OPEN mycursor FETCH NEXT FROM mycursor INTO @ordernum, @orderdate, @custnum, @repnum, @lastname, @firstname WHILE @@FETCH_STATUS = 0 BEGIN PRINT PRINT PRINT PRINT PRINT
@ordernum @orderdate @custnum @lastname @firstname
FETCH NEXT FROM mycursor INTO @ordernum, @orderdate, @custnum, @repnum, @lastname, @firstname END CLOSE mycursor DEALLOCATE mycursor Chapter 8
USING SQL IN MICROSOFT ACCESS Not every programming language accepts SQL commands as readily as PL/SQL and T-SQL. In Microsoft Access, programs are written in Visual Basic, which does not support embedded SQL commands directly in the code. When the SQL command is stored in a string variable, however, you can use the DoCmd.RunSQL command to run the command. The procedure in which you place the SQL command can include arguments.
Deleting Data with Visual Basic To delete the sales rep whose number is 20, the command is: DELETE FROM REP WHERE REP_NUM = '20';
259
When you write this type of command, you usually don’t know in advance the specific sales rep number that you want to delete; it would be passed as an argument to the procedure containing this DELETE command. In the following example, the sales rep number is stored in an argument named I_REP_NUM.
EXAMPLE 12 Delete from the REP table the sales rep whose number currently is stored in I_REP_NUM.
Statements in the procedure usually create the appropriate DELETE command, using the value in any necessary arguments. For example, when the command is stored in the variable named strSQL (which must be a string variable) and the rep number is stored in the argument I_REP_NUM, the following command is appropriate: strSQL = 'DELETE FROM REP WHERE REP_NUM = ' strSQL = strSQL & I_REP_NUM strSQL = strSQL & ';'
The first command sets the strSQL string variable to DELETE FROM REP WHERE REP_NUM = '; that is, it creates everything necessary in the command up to and including the single quotation mark preceding the rep number. The second command uses concatenation (&). It changes strSQL to the result of the previous value concatenated with the value in I_REP_NUM. When I_REP_NUM contains the value 20, for example, the command would be DELETE FROM REP WHERE REP_NUM = '20. The final command sets strSQL to the result of the value already created, concatenated with a single quotation mark and a semicolon. The command is now complete. Figure 8-27 shows a completed procedure to accomplish the necessary deletion in Access. You enter this procedure in the Microsoft Visual Basic window. In the program, the Dim statement creates a string variable named strSQL. The next three commands set strSQL to the appropriate SQL command. Finally, the DoCmd.RunSQL command runs the SQL command stored in strSQL.
SQL Functions and Procedures
Argument (I_REP_NUM)
Commands to set strSQL
260 Runs command stored in strSQL
FIGURE 8-27
Visual Basic code to delete a sales rep
NOTE If you have concerns about how you constructed the SQL command in strSQL, you can include the Debug.Print (strSQL) command after the set of commands that construct strSQL. The Debug.Print command displays the entire command before it is executed so you can review it for accuracy. If you need to correct an error, rerun the program after making the necessary changes. If you get an error in your program, check your SQL command carefully to make sure that you concatenated it correctly.
Running the Code Normally, you run code like the function shown in Figure 8-27 by calling it from another procedure or associating it with some event, such as clicking a button on a form. However, you can run it directly by using the Immediate window (click View on the menu bar, and then click Immediate Window to open it). Normally, you would use this window only for testing purposes, but you can use it to see the result of running the code. To run a Function procedure, such as the one shown in Figure 8-27, in the Immediate window, type a question mark followed by the name of the procedure and a set of parentheses, as shown in Figure 8-28. Place the values for any arguments in the parentheses. Assuming that you wanted to delete a sales rep whose number is 50, you would include '50' inside the parentheses as shown in the figure.
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261
Procedure to delete records in Immediate window
FIGURE 8-28
Running the code in the Immediate window
After you type the command and press the Enter key, the code will run and the appropriate action will occur. In this case, the command deletes the sales rep with the number 50 (assuming there is a sales rep 50).
Updating Data with Visual Basic A procedure that updates a table using an UPDATE command is similar to the one used to delete a sales rep. In Example 13, two arguments are required. One of them, I_LAST_NAME, contains the new name for the sales rep. The other, I_REP_NUM, contains the number of the rep whose name is to be changed.
EXAMPLE 13 Change the last name of the sales rep whose number is stored in I_REP_NUM to the value currently stored in I_LAST_NAME.
This example is similar to the previous one with two important differences. First, you need to use the UPDATE command instead of the DELETE command. Second, there are two arguments, so there are two portions of the construction of the SQL command that involve variables. The complete procedure is shown in Figure 8-29.
SQL Functions and Procedures
Public Function RepUpdate(I_LAST_NAME, I_REP_NUM) Dim strSQL As String strSQL strSQL strSQL strSQL strSQL
= = = = =
'UPDATE REP SET LAST_NAME = ' strSQL + I_LAST_NAME strSQL + ' WHERE REP_NUM = ' strSQL & I_REP_NUM strSQL & ';'
Argument giving number of rep whose name is to be changed
Argument giving new name
DoCmd.RunSQL strSQL End Function
262
FIGURE 8-29
Code to change a rep’s last name
To run this procedure, you would enter values for both arguments as shown in Figure 8-30.
New last name ?RepUpdate('Webb','20')
Rep number
FIGURE 8-30
Running the code to change a rep’s last name
Inserting Data with Visual Basic The process for inserting rows is similar in Access when compared to PL/SQL or T-SQL. You create the appropriate INSERT command in the strSQL variable. There will be multiple arguments in the procedure—one for each value to be inserted.
Finding Multiple Rows with Visual Basic Just as when embedding SQL in PL/SQL, deleting or updating multiple rows causes no problems, because these procedures still represent a single operation, with all the work happening behind the scenes. A SELECT command that returns several rows, however, poses serious problems for record-at-a-time languages like PL/SQL and Visual Basic. You handle SELECT commands differently in Access than you do in PL/SQL or T-SQL. In particular, there are no cursors in Access. Instead, you handle the results of a query just as you might use a loop to process through the records in a table.
EXAMPLE 14 Retrieve and list the number and name of each customer represented by the sales rep whose number is stored in the variable I_REP_NUM.
Chapter 8
Figure 8-31 shows a procedure to accomplish the indicated task. The statements involving rs and cnn are a typical way of processing through a recordset, that is, through all the records contained in a table or in the results of a query. The only difference between this program and one to process all the records in a table is that the Open command refers to an SQL command and not a table. (The SQL command is stored in the variable named strSQL and is created in the same manner as shown in the previous examples.)
Opens a recordset named rs for the query stored in strSQL
Public Function FindCustomers(I_REP_NUM) Dim rs As New ADODB.Recordset Dim cnn As ADODB.Connection Dim strSQL As String Set cnn = CurrentProject.Connection
Loops until end of file for recordset
rs.Open strSQL, cnn, adOpenStatic, , adCmdText Do Until rs.EOF Prints the customer Debug.Print (rs!CUSTOMER_NAME) name for the current rs.MoveNext record in recordset Loop
End of loop
End Function
FIGURE 8-31
263
strSQL = 'SELECT CUSTOMER_NAME FROM CUSTOMER WHERE REP_NUM = ' strSQL = strSQL & I_REP_NUM strSQL = strSQL & ';'
Moves to the next record in recordset
Code to find customers of a specific rep
The loop continues until reaching the end of file for the recordset, that is, until all records have been processed. Within the loop, you can use the Debug.Print command to print a value. In this case, the value to be printed is rs!CUSTOMER_NAME. This indicates the contents of the CUSTOMER_NAME column for the record in the recordset (rs) on which Access is currently positioned. The next command, rs.MoveNext, moves to the next record in the recordset. The loop continues until all records in the recordset have been processed. Figure 8-32 shows the results of running this procedure and entering a value of '35' as an argument. Access displays the four customers of sales rep 35.
Rep number ?FindCustomers('35') Brookings Direct The Everything Shop Lee's Sport and Appliance Deerfield's Four Seasons
FIGURE 8-32
Customers of the sales rep
Running the code to find customers of a sales rep
NOTE When you expect an SQL query to return only one record, you use the same process but would not need a loop.
SQL Functions and Procedures
U S I N G A TR I G G E R
264
A trigger is a procedure that is executed automatically in response to an associated database operation, such as an INSERT, UPDATE, or DELETE command. Unlike a stored procedure, which is executed in response to a user request, a trigger is executed in response to a command that causes the associated database operation to occur. The examples in this section assume there is a new column named ON_ORDER in the PART table. This column represents the number of units of a part currently on order. For example, if there are two separate order lines for a part and the number ordered on one order line is 3 and the number ordered on the other order line is 2, the ON_ORDER value for that part will be 5. Adding, changing, or deleting order lines affects the value in the ON_ORDER column for the part. To ensure that the value is updated appropriately, you can use a trigger. If you created the ADD_ORDER_LINE trigger shown in Figure 8-33, the SQL command in the trigger would be executed when a user adds an order line. The trigger must update the ON_ORDER value for the corresponding part to reflect the order line. For example, if the value in the ON_ORDER column for part CD52 is 4 and the user adds an order line on which the part number is CD52 and the number of units ordered is 2, six units of part CD52 will be on order. When a record is added to the ORDER_LINE table, the ADD_ORDER_LINE trigger updates the PART table by adding the number of units ordered on the order line to the previous value in the ON_ORDER column.
FIGURE 8-33
ADD_ORDER_LINE trigger
The first line indicates that the command is creating a trigger named ADD_ORDER_LINE. The second line indicates that this trigger will be executed after an order line is inserted and that the SQL command is to occur for each row that is added. Like stored procedures, the SQL command is enclosed between the words BEGIN and END. In this case, the SQL command is an UPDATE command. The command uses the NEW qualifier, which refers to the row that is added to the ORDER_LINE table. If an order line is added on which the part number is CD52 and the number ordered is 2, for example, NEW.PART_NUM will be CD52 and NEW.NUM_ORDERED will be 2. The following UPDATE_ORDER_LINE trigger shown in Figure 8-34 is executed when a user attempts to update an order line. There are two differences between the UPDATE_ORDER_LINE trigger and the ADD_ORDER_LINE trigger. First, the second line of the UPDATE_ORDER_LINE trigger indicates that this trigger is executed after an UPDATE of an order line rather than an INSERT. Second, the computation to update the ON_ORDER column includes both NEW.NUM_ORDERED and OLD.NUM_ORDERED. As with the ADD_ORDER_LINE trigger, NEW.NUM_ORDERED refers to the new value. In an
Chapter 8
UPDATE command, however, there is also an old value, which is the value before the update takes place. If an update changes the value for ON_ORDER from 1 to 3, OLD.NUM_ORDERED is 1 and NEW.NUM_ORDERED is 3. Adding NEW.NUM_ORDERED and subtracting OLD.NUM_ORDERED results in a net change of an increase of 2. (The net change could also be negative, in which case the ON_ORDER value decreases.)
265
FIGURE 8-34
UPDATE_ORDER_LINE trigger
The DELETE_ORDER_LINE trigger shown in Figure 8-35 performs a function similar to the other two. When an order line is deleted, the ON_ORDER value for the corresponding part is updated by subtracting OLD.NUM_ORDERED from the current ON_ORDER value. (In a delete operation, there is no NEW.NUM_ORDERED.)
FIGURE 8-35
DELETE_ORDER_LINE trigger
ACCESS USER NOTE Access does not support triggers. When using a form to update table data, you can achieve some of the same functionality by creating VBA code to be executed after the insertion, update, or deletion of records.
SQL Functions and Procedures
SQL SERVER USER NOTE In SQL Server, you create triggers using T-SQL. The code to create the ADD_ORDER_LINE trigger is: CREATE TRIGGER ADD_ORDER_LINE ON ORDER_LINE AFTER INSERT AS DECLARE @numbord decimal(3,0) SELECT @numbord = (SELECT NUM_ORDERED FROM INSERTED) UPDATE PART SET ON_ORDER = ON_ORDER + @numbord 266
This trigger uses one variable, @numbord, and the value placed in that variable is obtained from the SELECT statement. The INSERTED table is a temporary system table that contains a copy of the values that the last SQL command inserted. The column names are the same column names as in the ORDER_LINE table. The INSERTED table holds the most recent value of the NUM_ORDERED column which is what you need to update the PART table. The T-SQL trigger that executes after an UPDATE of an order line is: CREATE TRIGGER UPDATE_ORDER_LINE ON ORDER_LINE AFTER UPDATE AS DECLARE @newnumbord decimal(3,0) DECLARE @oldnumbord decimal(3,0) SELECT @newnumbord = (SELECT NUM_ORDERED FROM INSERTED) SELECT @oldnumbord = (SELECT NUM_ORDERED FROM DELETED) UPDATE PART SET ON_ORDER = ON_ORDER + @newnumbord - @oldnumbord This trigger uses the INSERTED table and the DELETED table. The DELETED table contains the previous value of the NUM_ORDERED column while the INSERTED column contains the updated value. The DELETE_ORDER_LINE trigger uses only the DELETED system table: CREATE TRIGGER DELETE_ORDER_LINE ON ORDER_LINE AFTER DELETE AS DECLARE @numbord decimal(3,0) SELECT @numbord = (SELECT NUM_ORDERED FROM DELETED) UPDATE PART SET ON_0RDER = ON_ORDER - @numbord
Chapter 8
Chapter Summary ●
There are functions whose results are based on the values in single records. UPPER and LOWER are two examples of functions that act on character data. UPPER displays each lowercase letter in the argument in uppercase. LOWER displays each uppercase letter in the argument in lowercase.
●
ROUND and FLOOR are two examples of functions that act on numeric data. ROUND produces its result by rounding the value to the specified number of decimal places. FLOOR produces its result by truncating (removing) everything to the right of the decimal point.
●
Use the ADD_MONTHS function in Oracle to add a specific number of months to a date. In Access and in SQL Server, use the DATEADD() function.
●
To add a specific number of days to a date, use normal addition. You can subtract one date from another to produce the number of days between two dates.
●
To obtain today’s date, use the SYSDATE function in Oracle, the GETDATE() function in SQL Server, and the DATE() function in Access.
●
To concatenate values in character columns in Oracle, separate the column names with two vertical lines (||). Use the RTRIM function to delete any extra spaces that follow the values. In SQL Server, use the + symbol to concatenate values. In Access, use the ampersand (&) symbol to concatenate values.
●
A stored procedure is a query saved in a file that users can execute later.
●
To create a stored procedure in PL/SQL or T-SQL, use the CREATE PROCEDURE command.
●
Variables in PL/SQL procedures are declared after the word DECLARE. To assign variables the same type as a column in the database, use the %TYPE attribute.
●
Use the INTO clause in the SELECT command to place the results of a SELECT command in variables in Oracle.
●
You can use INSERT, UPDATE, and DELETE commands in PL/SQL and T-SQL procedures, even when they affect more than one row.
●
When a SELECT command is used to retrieve more than one row in PL/SQL or T-SQL, it must define a cursor that will select one row at a time.
●
Use the OPEN command to activate a cursor and execute the query in the cursor definition.
●
Use the FETCH command to select the next row in PL/SQL and T-SQL.
●
Use the CLOSE command to deactivate a cursor. The rows initially retrieved will no longer be available to PL/SQL or T-SQL.
●
To use SQL commands in Access, create the command in a string variable. To run the command stored in the string variable, use the DoCmd.RunSQL command.
●
To process a collection of rows retrieved by a SELECT command in Access, use a recordset. Create the SQL command in a string variable and use the string variable in the command to open the recordset.
●
To move to the next record in a recordset in Access, use the MoveNext command.
267
SQL Functions and Procedures
●
A trigger is an action that occurs automatically in response to an associated database operation, such as an INSERT, UPDATE, or DELETE command. Like a stored procedure, a trigger is stored and compiled on the server. Unlike a stored procedure, which is executed in response to a user request, a trigger is executed in response to a command that causes the associated database operation to occur.
Key Terms
268
ADD_MONTHS
OPEN
argument
PL/SQL
call
procedural code
client
procedural language
client/server system
ROUND
CLOSE
RTRIM
concatenate
server
concatenation
stored procedure
cursor
SYSDATE
embed
Transact-SQL
FETCH
trigger
FLOOR
T-SQL
LOWER
update procedure
nonprocedural language
UPPER
Review Questions
Chapter 8
1.
How do you display letters in uppercase in Oracle, Access, and SQL Server? How do you display letters in lowercase in Oracle, Access, and SQL Server?
2.
How do you round a number to a specific number of decimal places in Oracle, Access, and SQL Server? How do you remove everything to the right of the decimal place in Oracle and SQL Server?
3.
How do you add months to a date in Oracle, Access, and SQL Server? How do you add days to a date? How would you find the number of days between two dates?
4.
How do you obtain today’s date in Oracle, Access, and SQL Server?
5.
How do you concatenate values in character columns in Oracle, Access, and SQL Server?
6.
Which function deletes extra spaces at the end of a value?
7.
What are stored procedures? What purpose do they serve?
8.
In which portion of a PL/SQL procedure do you embed SQL commands?
9.
Where do you declare variables in PL/SQL procedures?
10.
In PL/SQL, how do you assign variables the same type as a column in the database?
11.
How do you place the results of a SELECT command into variables in PL/SQL?
12.
Can you use INSERT, UPDATE, or DELETE commands that affect more than one row in PL/SQL procedures?
13.
How do you use a SELECT command that retrieves more than one row in a PL/SQL procedure?
14.
Which PL/SQL command activates a cursor?
15.
Which PL/SQL command selects the next row in a cursor?
16.
Which PL/SQL command deactivates a cursor?
17.
How do you use SQL commands in Access?
18.
How do you process a collection of rows retrieved by a SELECT command in Access?
19.
How do you move to the next record in a recordset in Access?
20.
What are triggers? What purpose do they serve?
21.
What is the purpose of the INSERTED and DELETED tables in SQL Server?
269
Exercises Premiere Products Use the Premiere Products database (see Figure 1-2 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
List the part number and description for all parts. The part descriptions should appear in uppercase letters.
2.
List the customer number and name for all customers located in the city of Grove.Your query should ignore case. For example, a customer with the city Grove should be included as should customers whose city is GROVE, grove, GrOvE, and so on.
3.
List the customer number, name, and balance for all customers. The balance should be rounded to the nearest dollar.
4.
Premiere Products is running a promotion that is valid for up to 20 days after an order is placed. List the order number, customer number, customer name, and the promotion date for each order. The promotion date is 20 days after the order was placed.
5.
Write PL/SQL or T-SQL procedures to accomplish the following tasks: a.
Obtain the name and credit limit of the customer whose number currently is stored in I_CUSTOMER_NUM. Place these values in the variables I_CUSTOMER_NAME and I_CREDIT_LIMIT, respectively. Output the contents of I_CUSTOMER_NAME and I_CREDIT_LIMIT.
b.
Obtain the order date, customer number, and name for the order whose number currently is stored in I_ORDER_NUM. Place these values in the variables I_ORDER_DATE, I_CUSTOMER_NUM and I_CUSTOMER_NAME, respectively. Output the contents of I_ORDER_DATE, I_CUSTOMER_NUM, and I_CUSTOMER_NAME.
c.
Add a row to the ORDERS table.
SQL Functions and Procedures
d.
Change the date of the order whose number is stored in I_ORDER_NUM to the date currently found in I_ORDER_DATE.
e.
Delete the order whose number is stored in I_ORDER_NUM.
6.
Write a PL/SQL or T-SQL procedure to retrieve and output the part number, part description, warehouse number, and unit price of every part in the item class stored in I_CLASS.
7.
Write Access functions to accomplish the following tasks:
270
a.
Delete the order whose number is stored in I_ORDER_NUM.
b.
Change the date of the order whose number is stored in I_ORDER_NUM to the date currently found in I_ORDER_DATE.
c.
Retrieve and output the part number, part description, warehouse number, and unit price of every part in the item class stored in I_CLASS.
8.
Write a stored procedure in PL/SQL or T-SQL that will change the price of a part with a given part number. How would you use this stored procedure to change the price of part AT94 to $26.95?
9.
Write the code for the following triggers in PL/SQL or T-SQL following the style shown in the text: a.
When adding a customer, add the customer’s balance times the sales rep’s commission rate to the commission for the corresponding sales rep.
b.
When updating a customer, add the difference between the new balance and the old balance multipled by the sales rep’s commission rate to the commission for the corresponding sales rep.
c.
When deleting a customer, subtract the balance multiplied by the sales rep’s commission rate from the commission for the corresponding sales rep.
Henry Books Use the Henry Books database (see Figures 1-4 through 1-7 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
List the author number, first name, and last name for all authors. The first name should appear in lowercase letters and the last name should appear in uppercase letters.
2.
List the publisher code and name for all publishers located in the city of New York.Your query should ignore case. For example, a customer with the city New York should be included as should customers whose city is NEW YORK, New york, NeW yOrK, and so on.
3.
List the book code, title, and price for all books. The price should be rounded to the nearest dollar.
4.
Write PL/SQL or T-SQL procedures to accomplish the following tasks: a.
Chapter 8
Obtain the first name and last name of the author whose number currently is stored in I_AUTHOR_NUM. Place these values in the variables I_AUTHOR_FIRST and I_AUTHOR_LAST. Output the contents of I_AUTHOR_NUM, I_AUTHOR_FIRST, and I_AUTHOR_LAST.
b.
Obtain the book title, publisher code, and publisher name for every book whose code currently is stored in I_BOOK_CODE. Place these values in the variables I_TITLE, I_PUBLISHER_CODE, and I_PUBLISHER_NAME, respectively. Output the contents of I_TITLE, I_PUBLISHER_CODE, and I_PUBLISHER_NAME.
c.
Add a row to the AUTHOR table.
d.
Change the last name of the author whose number is stored in I_AUTHOR_NUM to the value currently found in I_AUTHOR_LAST.
e.
Delete the author whose number is stored in I_AUTHOR_NUM.
5.
Write a PL/SQL or T-SQL procedure to retrieve and output the book code, title, book type, and price for every book whose publisher code is stored in I_PUBLISHER_CODE.
6.
Write Access functions to accomplish the following tasks:
271
a.
Delete the author whose number is stored in I_AUTHOR_NUM.
b.
Change the last name of the author whose number is stored in I_AUTHOR_NUM to the value currently found in I_AUTHOR_LAST.
c.
Retrieve and output the book code, title, book type, and price for every book whose publisher code is stored in I_PUBLISHER_CODE.
7.
Write a stored procedure in PL/SQL or T-SQL that will change the price of a book with a given book code. How would you use this stored procedure to change the price of book 0189 to $8.49?
8.
Assume the BOOK table contains a column called TOTAL_ON_HAND that represents the total units on hand in all branches for that book. Following the style shown in the text, write the code in PL/SQL or T-SQL for the following triggers: a.
When inserting a row in the INVENTORY table, add the ON_HAND value to the TOTAL_ON_HAND value for the appropriate book.
b.
When updating a row in the INVENTORY table, add the difference between the new ON_HAND value and the old ON_HAND value to the TOTAL_ON_HAND value for the appropriate book.
c.
When deleting a row in the INVENTORY table, subtract the ON_HAND value from the TOTAL_ON_HAND value for the appropriate book.
Alexamara Marina Group Use the Alexamara Marina Group database (see Figures 1-8 through 1-12 in Chapter 1) to complete the following exercises. If directed to do so by your instructor, use the information provided with the Chapter 3 Exercises to print your output. 1.
List the owner number, first name, and last name for all owners. The first name should appear in uppercase letters and the last name should appear in lowercase letters.
2.
List the owner number and last name for all owners located in the city of Bowton. Your query should ignore case. For example, a customer with the city Bowton should be included as should customers whose city is BOWTON, BowTon, BoWtOn, and so on.
SQL Functions and Procedures
3.
Alexamara is offering a discount for owners who sign up early for slips for next year. The discount is 1.75 percent of the rental fee. For each slip, list the marina number, slip number, owner number, owner’s last name, rental fee, and discount. The discount should be rounded to the nearest dollar.
4.
Write PL/SQL or T-SQL procedures to accomplish the following tasks: a.
Obtain the first name and last name of the owner whose number currently is stored in I_OWNER_NUM. Place these values in the variables I_FIRST_NAME and I_LAST_NAME. Output the contents of I_OWNER_NUM, I_FIRST_NAME, and I_LAST_NAME.
b.
Obtain the marina number, slip number, boat name, owner number, owner first name, and owner last name for the slip whose slip ID is currently stored in I_SLIP_ID. Place these values in the variables I_MARINA_NUM, I_SLIP_NUM, I_BOAT_NAME, I_OWNER_NUM, I_FIRST_NAME, and I_LAST_NAME, respectively. Output the contents of I_SLIP_ID, I_MARINA_NUM, I_SLIP_NUM, I_BOAT_NAME, I_OWNER_NUM, I_FIRST_NAME, and I_LAST_NAME.
c.
Add a row to the OWNER table.
d.
Change the last name of the owner whose number is stored in I_OWNER_NUM to the value currently found in I_LAST_NAME.
e.
Delete the owner whose number is stored in I_OWNER_NUM.
272
Chapter 8
5.
Write a PL/SQL or T-SQL procedure to retrieve and output the marina number, slip number, rental fee, boat name, and owner number for every slip whose length is equal to the length stored in I_LENGTH.
6.
Write Access functions to accomplish the following tasks: a.
Delete the owner whose number is stored in I_OWNER_NUM.
b.
Change the last name of the owner whose number is stored in I_OWNER_NUM to the value currently found in I_LAST_NAME.
c.
Retrieve and output the marina number, slip number, rental fee, boat name, and owner number for every slip whose length is equal to the length stored in I_LENGTH.
7.
Write a stored procedure in PL/SQL or T-SQL that will change the rental fee of a slip with a given slip ID and marina number. How would you use this stored procedure to change the rental fee for the boat with the slip ID 3 in marina 1 to $3,700?
8.
Assume the OWNER table contains a column called TOTAL_RENTAL that represents the total rental fee for all slips rented by that owner. Write the code in PL/SQL or T-SQL for the following triggers following the style shown in the text: a.
When inserting a row in the MARINA_SLIP table, add the rental fee to the total rental for the appropriate owner.
b.
When updating a row in the MARINA_SLIP table, add the difference between the new rental fee and the old rental fee to the total rental for the appropriate owner.
c.
When deleting a row in the MARINA_SLIP table, subtract the rental fee from the total rental for the appropriate owner.
APPENDIX
A
S Q L R EF EREN C E INTRODUCTION You can use this appendix to obtain details concerning important components and syntax of the SQL language. Items are arranged alphabetically. Each item contains a description, a reference to where the item is covered in the text, and, when appropriate, both an example and a description of the query results. Some SQL commands also include a description of the clauses associated with them. For each clause, there is a brief description and an indication of whether the clause is required or optional.
ALIASES (PAGES 146–148) You can specify an alias (alternative name) for each table in a query. You can use the alias in the rest of the command by following the name of the table with a space and the alias name. The following command creates an alias named R for the REP table and an alias named C for the CUSTOMER table: SELECT R.REP_NUM, R.LAST_NAME, R.FIRST_NAME, C.CUSTOMER_NUM, C.CUSTOMER_NAME FROM REP R, CUSTOMER C WHERE R.REP_NUM = C.REP_NUM;
A LT E R TA B L E ( PA G E S 1 8 2 – 1 8 3 ) Use the ALTER TABLE command to change a table’s structure. As shown in Figure A-1, you type the ALTER TABLE command, followed by the table name, and then the alteration to perform.
Clause
Description
Required?
ALTER TABLE table name
Indicates name of table to be altered.
Yes
alteration
Indicates type of alteration to be performed.
Yes
FIGURE A-1
ALTER TABLE command
274
The following command alters the CUSTOMER table by adding a new CUSTOMER_TYPE column: ALTER TABLE CUSTOMER ADD CUSTOMER_TYPE CHAR(1);
The following command changes the CITY column in the CUSTOMER table so that it cannot accept nulls: ALTER TABLE CUSTOMER MODIFY CITY NOT NULL;
Note: In Access, you usually make these changes to a table in Design view rather than using ALTER TABLE. Note: In SQL Server, you must use the ALTER COLUMN clause and completely define the column as follows: ALTER TABLE CUSTOMER ALTER COLUMN CITY CHAR(15) NOT NULL
COLUMN OR EXPRESSION LIST (SELECT CLAUSE) (PAGES 98–100) To select columns, use the SELECT clause followed by the list of columns, separated by commas. The following SELECT clause selects the CUSTOMER_NUM, CUSTOMER_NAME, and BALANCE columns: SELECT CUSTOMER_NUM, CUSTOMER_NAME, BALANCE
Use an asterisk in a SELECT clause to select all columns in a table. The following SELECT clause selects all columns: SELECT *
Computed Columns (Pages 107–110) You can use a computation in place of a column by typing the computation. For readability, you can type the computation in parentheses, although it is not necessary to do so. The following SELECT clause selects the CUSTOMER_NUM and CUSTOMER_NAME columns as well as the results of subtracting the BALANCE column from the CREDIT_LIMIT column: SELECT CUSTOMER_NUM, CUSTOMER_NAME, (CREDIT_LIMIT - BALANCE)
The DISTINCT Operator (Pages 117–119) To avoid selecting duplicate values in a command, use the DISTINCT operator. When you omit the DISTINCT operator from the command and the same value appears on multiple rows in the table, that value will appear on multiple rows in the query results. The following query selects all customer numbers from the ORDERS table, but lists each customer number only once in the results: SELECT DISTINCT(CUSTOMER_NUM) FROM ORDERS; Appendix A
Functions (Pages 114–117)
275
You can use functions in a SELECT clause. The most commonly used functions are AVG (to calculate an average), COUNT (to count the number of rows), MAX (to determine the maximum value), MIN (to determine the minimum value), and SUM (to calculate a total). The following SELECT clause calculates the average balance: SELECT AVG(BALANCE)
COMMIT (PAGES 177–178) Use the COMMIT command to make permanent any updates made since the last command. If no previous COMMIT command has been executed, the COMMIT command will make all the updates during the current work session permanent immediately. All updates become permanent automatically when you exit SQL. Figure A-2 describes the COMMIT command.
Clause
Description
Required?
COMMIT
Indicates that a COMMIT is to be performed.
Yes
FIGURE A-2
COMMIT command
The following command makes all updates since the most recent COMMIT command permanent: COMMIT;
Note: In SQL Server, the following command makes all updates since the most recent COMMIT command: COMMIT TRANSACTION
Note: Access does not support the COMMIT command.
CONDITIONS (PAGES 100–106) A condition is an expression that can be evaluated as either true or false. When you use a condition in a WHERE clause, the results of the query contain those rows for which the condition is true. You can create simple conditions and compound conditions using the BETWEEN, LIKE, IN, EXISTS, ALL, and ANY operators, as described in the following sections.
Simple Conditions (Pages 100–103) A simple condition has the form column name, comparison operator, and then either another column name or a value. The available comparison operators are = (equal to), < (less than), > (greater than), <= (less than or equal to), >= (greater than or equal to), and < > (not equal to). The following WHERE clause uses a condition to select rows where the balance is greater than the credit limit: WHERE BALANCE > CREDIT_LIMIT
SQL Reference
276
Compound Conditions (Pages 103–106) Compound conditions are formed by connecting two or more simple conditions using the AND, OR, and NOT operators. When simple conditions are connected by the AND operator, all of the simple conditions must be true in order for the compound condition to be true. When simple conditions are connected by the OR operator, the compound condition will be true whenever any one of the simple conditions is true. Preceding a condition by the NOT operator reverses the truth of the original condition. The following WHERE clause is true if the warehouse number is equal to 3 or the units on hand is greater than 100, or both: WHERE (WAREHOUSE = '3') OR (ON_HAND > 100)
The following WHERE clause is true if the warehouse number is equal to 3 and the units on hand is greater than 100: WHERE (WAREHOUSE = '3') AND (ON_HAND > 100)
The following WHERE clause is true if the warehouse number is not equal to 3: WHERE NOT (WAREHOUSE = '3')
BETWEEN Conditions (Pages 106–107) You can use the BETWEEN operator to determine if a value is within a range of values. The following WHERE clause is true if the balance is between 2,000 and 5,000: WHERE BALANCE BETWEEN 2000 AND 5000
LIKE Conditions (Pages 110–111) LIKE conditions use wildcards to select rows. Use the percent (%) wildcard to represent any collection of characters. The condition LIKE '%Central%' will be true for data consisting of any character or characters, followed by the letters “Central,” followed by any other character or characters. Another wildcard symbol is the underscore (_), which represents any individual character. For example, “T_m” represents the letter “T,” followed by any single character, followed by the letter “m,” and would be true for a collection of characters such as Tim, Tom, or T3m. The following WHERE clause is true if the value in the STREET column is Central, Centralia, or any other value that contains “Central”: WHERE STREET LIKE '%Central%'
Note: Access uses different wildcard symbols. The symbol for any collection of characters is the asterisk (*). The symbol for an individual character is the question mark (?).
IN Conditions (Pages 111–112, 140–141) You can use IN to determine whether a value is in some specific collection of values. The following WHERE clause is true if the credit limit is 5,000, 10,000, or 15,000: WHERE CREDIT_LIMIT IN (5000, 10000, 15000)
Appendix A
The following WHERE clause is true if the part number is in the collection of part numbers associated with order number 21610:
277
WHERE PART_NUM IN (SELECT PART_NUM FROM ORDER_LINE WHERE ORDER_NUM = '21610')
EXISTS Conditions (Pages 141–142) You can use EXISTS to determine whether the results of a subquery contain at least one row. The following WHERE clause is true if the results of the subquery contain at least one row, that is, there is at least one order line with the desired order number and on which the part number is DR93: WHERE EXISTS (SELECT * FROM ORDER_LINE WHERE ORDERS.ORDER_NUM = ORDER_LINE.ORDER_NUM AND PART_NUM = 'DR93')
ALL and ANY (Pages 157–160) You can use ALL or ANY with subqueries. If you precede the subquery by ALL, the condition is true only if it is satisfied for all values produced by the subquery. If you precede the subquery by ANY, the condition is true if it is satisfied for any value (one or more) produced by the subquery. The following WHERE clause is true if the balance is greater than every balance contained in the results of the subquery: WHERE BALANCE > ALL (SELECT BALANCE FROM CUSTOMER WHERE REP_NUM = '65')
The following WHERE clause is true if the balance is greater than at least one balance contained in the results of the subquery: WHERE BALANCE > ANY (SELECT BALANCE FROM CUSTOMER WHERE REP_NUM = '65')
C R E AT E I N D E X ( PA G E S 2 1 5 – 2 1 7 ) Use the CREATE INDEX command to create an index for a table. Figure A-3 describes the CREATE INDEX command.
SQL Reference
278
Clause
Description
Required?
CREATE INDEX index name
Indicates the name of the index.
Yes
ON table name
Indicates the table for which the index is to be created.
Yes
column list
Indicates the column or columns on which the index is to be based.
Yes
FIGURE A-3
CREATE INDEX command
The following CREATE INDEX command creates an index named REPNAME for the REP table on the combination of the LAST_NAME and FIRST_NAME columns: CREATE INDEX REPNAME ON REP(LAST_NAME, FIRST_NAME);
C R E AT E TA B L E ( PA G E S 6 6 – 6 9 ) Use the CREATE TABLE command to define the structure of a new table. Figure A-4 describes the CREATE TABLE command. Clause
Description
Required?
CREATE TABLE table name
Indicates the name of the table to be created.
Yes
(column and data type list)
Indicates the columns that comprise the table along with their corresponding data types (see Data Types section).
Yes
FIGURE A-4
CREATE TABLE command
The following CREATE TABLE command creates the REP table and its associated columns and data types. REP_NUM is the table’s primary key. CREATE TABLE REP (REP_NUM CHAR(2) PRIMARY KEY, LAST_NAME CHAR(15), FIRST_NAME CHAR(15), STREET CHAR(15), CITY CHAR(15), STATE CHAR(2), ZIP CHAR(5), COMMISSION DECIMAL(7,2), RATE DECIMAL(3,2) );
Appendix A
Note: Access does not support the DECIMAL data type. Use the CURRENCY data type for fields that will contain currency values; use the NUMBER data type for all other numeric fields. In Access, use the following command to create the REP table:
279
CREATE TABLE REP (REP_NUM CHAR(2) PRIMARY KEY, LAST_NAME CHAR(15), FIRST_NAME CHAR(15), STREET CHAR(15), CITY CHAR(15), STATE CHAR(2), ZIP CHAR(5), COMMISSION CURRENCY, RATE NUMBER );
C R E AT E V I E W ( PA G E S 1 9 6 – 1 9 7 ) Use the CREATE VIEW command to create a view. Figure A-5 describes the CREATE VIEW command.
Clause
Description
Required?
CREATE VIEW view name AS
Indicates the name of the view to be created.
Yes
query
Indicates the defining query for the view.
Yes
FIGURE A-5
CREATE VIEW command
The following CREATE VIEW command creates a view named HOUSEWARES, which consists of the part number, part description, units on hand, and unit price for all rows in the PART table on which the item class is HW: CREATE VIEW HOUSEWARES AS SELECT PART_NUM, PART_DESCRIPTION, ON_HAND, PRICE FROM PART WHERE CLASS = 'HW';
DATA TYPES (PAGE 71) Figure A-6 describes the data types that you can use in a CREATE TABLE command.
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Data Type
Description
CHAR(n)
Stores a character string n characters long. You use the CHAR type for columns that contain letters and special characters and for columns containing numbers that will not be used in any calculations. Because neither sales rep numbers nor customer numbers will be used in any calculations, for example, the REP_NUM and CUSTOMER_NUM columns are both assigned theCHAR data type.
VARCHAR(n)
An alternative to CHAR that stores a character string up to n characters long. Unlike CHAR, only the actual character string is stored. If a character string 20 characters long is stored in a CHAR(30) column, for example, it will occupy 30 characters (20 characters plus 10 blank spaces). If it is stored in a VARCHAR(30) column, it will only occupy 20 spaces. In general, tables that use VARCHAR instead of CHAR occupy less space, but the DBMS does not process them as rapidly during queries and updates. However, both are legitimate choices. This text uses CHAR, but VARCHAR would work equally well.
DATE
Stores date data. The specific format in which dates are stored varies from one SQL implementation to another. In Oracle, dates are enclosed in single quotation marks and have the form DD-MON-YYYY (for example,'15-OCT-2010' is October 15, 2010). In Access, dates are enclosed in number signs and are entered using the format MM/DD/YYYY (for example, #10/15/2010# is October 15, 2010). In SQL Server, use the DATETIME data type to store dates.
DECIMAL(p,q)
Stores a decimal number p digits long with q of these digits being decimal places to the right of the decimal point. For example, the data type DECIMAL(5,2) represents a number with three places to the left and two places to the right of the decimal (for example, 100.00). You can use the contents of DECIMAL columns in calculations. You also can use the NUMBER(p,q) data type in both Oracle and SQL Server to store a decimal number. Access does not support the DECIMAL data type; use the CURRENCY or NUMBER data type instead.
INT
Stores integers, which are numbers without a decimal part. The valid range is -2147483648 to 2147483647. You can use the contents of INT columns in calculations. If you follow the word INT with AUTO_INCREMENT, you create a column for which SQL will automatically generate a new sequence number each time you add a new row. This would be the appropriate choice, for example, when you want the DBMS to generate a value for a primary key.
SMALLINT
Stores integers, but uses less space than the INT data type. The valid range is -32768 to 32767. SMALLINT is a better choice than INT when you are certain that the column will store numbers within the indicated range. You can use the contents of SMALLINT columns in calculations.
FIGURE A-6
Data types
DELETE ROWS (PAGES 79, 178–180) Use the DELETE command to delete one or more rows from a table. Figure A-7 describes the DELETE command.
Clause
Description
Required?
DELETE table name
Indicates the table from which the row or rows are to be deleted.
Yes
WHERE condition
Indicates a condition. Those rows for which the condition is true will be retrieved and deleted.
No (If you omit the WHERE clause, all rows will be deleted.)
FIGURE A-7
Appendix A
DELETE command
The following DELETE command deletes any row from the LEVEL1_CUSTOMER table on which the customer number is 842:
281
DELETE LEVEL1_CUSTOMER WHERE CUSTOMER_NUM = '842';
DESCRIBE (PAGES 87–88) In Oracle, you can use the DESCRIBE command to list all the columns in a table and their properties. The following command describes the REP table: DESCRIBE REP;
Note: In Access, use the Documenter to describe the tables and other objects in a database. Note: In SQL Server, execute the sp_columns command to list all the columns in a table. The following command will list all the columns in the REP table: Exec sp_columns REP
DROP INDEX (PAGE 217) Use the DROP INDEX command to delete an index, as shown in Figure A-8.
Clause
Description
Required?
DROP INDEX index name
Indicates the name of the index to be dropped.
Yes
FIGURE A-8
DROP INDEX command
The following DROP INDEX command deletes the index named CREDNAME: DROP INDEX CREDNAME;
Note: In SQL Server, you must qualify the index name as follows: DROP INDEX CUSTOMER.CREDNAME
D R O P TA B L E ( PA G E S 7 0 , 1 8 9 ) Use the DROP TABLE command to delete a table, as shown in Figure A-9.
Clause
Description
Required?
DROP TABLE table name
Indicates name of the table to be dropped.
Yes
FIGURE A-9
DROP TABLE command
The following DROP TABLE command deletes the table named LEVEL1_CUSTOMER: DROP TABLE LEVEL1_CUSTOMER;
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DROP VIEW (PAGES 208–209) Use the DROP VIEW command to delete a view, as shown in Figure A-10.
Clause
Description
Required?
DROP VIEW view name
Indicates the name of the view to be dropped.
Yes
FIGURE A-10
DROP VIEW command
The following DROP VIEW command deletes the view named HSEWRES: DROP VIEW HSEWRES;
GRANT (PAGES 209–212) Use the GRANT command to grant privileges to a user. Figure A-11 describes the GRANT command.
Clause
Description
Required?
GRANT privilege
Indicates the type of privilege(s) to be granted.
Yes
ON database object
Indicates the database object(s) to which the privilege(s) pertain.
Yes
TO user name
Indicates the user(s) to whom the privilege(s) are to be granted. To grant the privilege(s) to all users, use the TO PUBLIC clause.
Yes
FIGURE A-11
GRANT command
The following GRANT command grants the user named Johnson the privilege of selecting rows from the REP table: GRANT SELECT ON REP TO Johnson;
INSERT INTO (QUERY) (PAGES 172–173) Use the INSERT INTO command with a query to insert the rows retrieved by a query into a table. As shown in Figure A-12, you must indicate the name of the table into which the row(s) will be inserted and the query whose results will be inserted into the named table.
Clause
Description
Required?
INSERT INTO table name
Indicates the name of the table into which the row(s) will be inserted.
Yes
query
Indicates the query whose results will be inserted into the table.
Yes
FIGURE A-12
Appendix A
INSERT INTO (query) command
The following INSERT INTO command inserts rows selected by a query into the LEVEL1_CUSTOMER table:
283
INSERT INTO LEVEL1_CUSTOMER SELECT CUSTOMER_NUM, CUSTOMER_NAME, BALANCE, CREDIT_LIMIT, REP_NUM FROM CUSTOMER WHERE CREDIT_LIMIT = 7500;
INSERT INTO (VALUES) (PAGES 72–75) Use the INSERT INTO command and the VALUES clause to insert a row into a table by specifying the values for each of the columns. As shown in Figure A-13, you must indicate the table into which to insert the values, and then list the values to insert in parentheses.
Clause
Description
Required?
INSERT INTO table name
Indicates the name of the table into which the row will be inserted.
Yes
VALUES (values list)
Indicates the values for each of the columns on the new row.
Yes
FIGURE A-13
INSERT INTO (values) command
The following INSERT INTO command inserts the values shown in parentheses as a new row in the REP table: INSERT INTO REP VALUES ('20','Kaiser','Valerie','624 Randall','Grove','FL','33321',20542.50,0.05);
INTEGRITY (PAGES 221–225) You can use the ALTER TABLE command with an appropriate ADD CHECK, ADD PRIMARY KEY, or ADD FOREIGN KEY clause to specify integrity. Figure A-14 describes the ALTER TABLE command for specifying integrity.
Clause
Description
Required?
ALTER TABLE table name
Indicates the table for which integrity is being specified.
Yes
integrity clause
ADD CHECK, ADD PRIMARY KEY, or ADD FOREIGN KEY
Yes
FIGURE A-14
Integrity options
The following ALTER TABLE command changes the PART table so that the only legal values for the CLASS column are AP, HW, and SG: ALTER TABLE PART ADD CHECK (CLASS IN ('AP','HW','SG') );
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The following ALTER TABLE command changes the REP table so that the REP_NUM column is the table’s primary key: ALTER TABLE REP ADD PRIMARY KEY(REP_NUM);
The following ALTER TABLE command changes the CUSTOMER table so that the REP_NUM column in the CUSTOMER table is a foreign key referencing the primary key of the REP table: ALTER TABLE CUSTOMER ADD FOREIGN KEY(REP_NUM) REFERENCES REP;
REVOKE (PAGES 209–212) Use the REVOKE command to revoke privileges from a user. Figure A-15 describes the REVOKE command. Clause
Description
REVOKE privilege
Indicates the type of privilege(s) to be revoked.
Yes
ON database object
Indicates the database object(s) to which the privilege pertains.
Yes
FROM user name
Indicates the user name(s) from whom the privilege(s) are to be revoked.
Yes
FIGURE A-15
Required?
REVOKE command
The following REVOKE command revokes the SELECT privilege for the REP table from the user named Johnson: REVOKE SELECT ON REP FROM Johnson;
ROLLBACK (PAGES 177–178) Use the ROLLBACK command to reverse (undo) all updates since the execution of the previous COMMIT command. If no COMMIT command has been executed, the command will undo all changes made during the current work session. Figure A-16 describes the ROLLBACK command.
Clause
Description
Required?
ROLLBACK
Indicates that a rollback is to be performed.
Yes
FIGURE A-16
ROLLBACK command
The following command reverses all updates made since the time of the last COMMIT command: ROLLBACK;
Appendix A
Note: In SQL Server, the following command reverses all updates made since the time of the last COMMIT command:
285
ROLLBACK TRANSACTION
Note: Access does not support the ROLLBACK command.
SELECT (PAGES 75–78, 98–127) Use the SELECT command to retrieve data from a table or from multiple tables. Figure A-17 describes the SELECT command.
Clause
Description
Required?
SELECT column or expression list
Indicates the column(s) and/or expression(s) to be retrieved.
Yes
FROM table list
Indicates the table(s) required for the query.
Yes
WHERE condition
Indicates one or more conditions. Only the rows for which the condition(s) are true will be retrieved.
No (If you omit the WHERE clause, all rows will be retrieved.)
GROUP BY column list
Indicates column(s) on which rows are to be grouped.
No (If you omit the GROUP BY clause, no grouping will occur.)
HAVING condition involving groups
Indicates a condition for groups. Only groups for which the condition is true will be included in query results. Use the HAVING clause only if the query output is grouped.
No (If you omit the HAVING clause, all groups will be included.)
ORDER BY column or expression list
Indicates column(s) on which the query output is to be sorted.
No (If you omit the ORDER BY clause, no sorting will occur.)
FIGURE A-17
SELECT command
The following SELECT command joins the ORDERS and ORDER_LINE tables. The command selects the customer number, order number, order date, and the sum of the product of the number ordered and unit price, renamed as ORDER_TOTAL. Records are grouped by order number, customer number, and order date. Only groups on which the order total is greater than 1,000 are included. Groups are ordered by order number. SELECT CUSTOMER_NUM, ORDERS.ORDER_NUM, ORDER_DATE, SUM(NUM_ORDERED * QUOTED_PRICE) AS ORDER_TOTAL FROM ORDERS, ORDER_LINE WHERE ORDERS.ORDER_NUM = ORDER_LINE.ORDER_NUM GROUP BY ORDERS.ORDER_NUM, CUSTOMER_NUM, ORDER_DATE HAVING SUM(NUM_ORDERED * QUOTED_PRICE) > 1000 ORDER BY ORDERS.ORDER_NUM;
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SUBQUERIES (PAGES 120–123, 142–144) You can use one query within another. The inner query is called a subquery and it is evaluated first. The outer query is evaluated next. The following command contains a subquery that produces a list of part numbers included in order number 21610: SELECT PART_DESCRIPTION FROM PART WHERE PART_NUM IN (SELECT PART_NUM FROM ORDER_LINE WHERE ORDER_NUM = '21610');
U N I O N , I N T E R S E C T, A N D M I N U S ( PA G E S 152–157) Connecting two SELECT commands with the UNION operator produces all the rows that would be in the results of the first query, the second query, or both queries. Connecting two SELECT commands with the INTERSECT operator produces all the rows that would be in the results of both queries. Connecting two SELECT commands with the MINUS operator produces all the rows that would be in the results of the first query, but not in the results of the second query. Figure A-18 describes the UNION, INTERSECT, and MINUS operators.
Operator
Description
UNION
Produces all the rows that would be in the results of the first query, the second query, or both queries.
INTERSECT
Produces all the rows that would be in the results of both queries.
MINUS
Produces all the rows that would be in the results of the first query but not in the results of the second query.
FIGURE A-18
UNION, INTERSECT, and MINUS operators
Note: Access and SQL Server support the UNION operation. SQL Server supports the INTERSECT command while Access does not. Neither SQL Server nor Microsoft Access support the MINUS operator. The following query displays the customer number and customer name of all customers that are represented by sales rep 65, or that have orders, or both: SELECT CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER WHERE REP_NUM = '65' UNION SELECT CUSTOMER.CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER, ORDERS WHERE CUSTOMER.CUSTOMER_NUM = ORDERS.CUSTOMER_NUM;
Appendix A
The following query displays the customer number and customer name of all customers that are represented by sales rep 65 and that have orders:
287
SELECT CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER WHERE REP_NUM = '65' INTERSECT SELECT CUSTOMER.CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER, ORDERS WHERE CUSTOMER.CUSTOMER_NUM = ORDERS.CUSTOMER_NUM;
The following query displays the customer number and customer name of all customers that are represented by sales rep 65 but that do not have orders: SELECT CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER WHERE REP_NUM = '65' MINUS SELECT CUSTOMER.CUSTOMER_NUM, CUSTOMER_NAME FROM CUSTOMER, ORDERS WHERE CUSTOMER.CUSTOMER_NUM = ORDERS.CUSTOMER_NUM;
U P DAT E ( PA G E S 7 8 – 8 0 , 1 7 3 – 1 7 5 ) Use the UPDATE command to change the contents of one or more rows in a table. Figure A-19 describes the UPDATE command.
Clause
Description
Required?
UPDATE table name
Indicates the table whose contents will be changed.
Yes
SET column = expression
Indicates the column to be changed, along with an expression that provides the new value.
Yes
WHERE condition
Indicates a condition. The change will occur only on those rows for which the condition is true.
No (If you omit the WHERE clause, all rows will be updated.)
FIGURE A-19
UPDATE command
The following UPDATE command changes the customer name on the row in LEVEL1_CUSTOMER on which the customer number is 842 to All Season Sport: UPDATE LEVEL1_CUSTOMER SET CUSTOMER_NAME = 'All Season Sport' WHERE CUSTOMER_NUM = '842';
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APPENDIX
B
H O W DO I REF EREN C E
This appendix answers frequently asked questions about how to accomplish a variety of tasks using SQL. Use the second column to locate the correct section in Appendix A that answers your question.
How Do I
Review the Named Section(s) in Appendix A
Add columns to an existing table?
ALTER TABLE
Add rows?
INSERT INTO (Values)
Calculate a statistic (sum, average, maximum, minimum, or count)?
1. SELECT 2. Column or Expression List (SELECT Clause) (Use the appropriate function in the query.)
Change rows?
UPDATE
Create a data type for a column?
1. Data Types 2. CREATE TABLE
Create a table?
CREATE TABLE
Create a view?
CREATE VIEW
Create an index?
CREATE INDEX
Describe a table’s layout?
DESCRIBE
Delete a table?
DROP TABLE
Delete a view?
DROP VIEW
Delete an index?
DROP INDEX
Delete rows?
DELETE Rows
Drop a table?
DROP TABLE
Drop a view?
DROP VIEW
Drop an index?
DROP INDEX
Grant a privilege?
GRANT
Group data in a query?
SELECT (Use a GROUP BY clause.)
Insert rows?
INSERT INTO (Values)
Insert rows using a query?
INSERT INTO (Query)
Join tables?
Conditions (Include a WHERE clause to relate the tables.)
Make updates permanent?
COMMIT
Order query results?
SELECT (Use the ORDER BY clause.)
Prohibit nulls?
1. CREATE TABLE 2. ALTER TABLE (Include the NOT NULL clause in a CREATE TABLE or ALTER TABLE command.)
Remove a privilege?
REVOKE
Remove rows?
DELETE Rows
Retrieve all columns?
1. SELECT 2. Column or Expression List (SELECT Clause) (Type *in the SELECT clause.)
290
FIGURE B-1
Appendix B
How Do I reference
How Do I
Review the Named Section(s) in Appendix A
Retrieve all rows?
SELECT (Omit the WHERE clause.)
291
Retrieve only certain columns?
1. SELECT 2. Column or Expression List (SELECT Clause) (Type the list of columns in the SELECT clause.)
Revoke a privilege?
REVOKE
Select all columns?
1. SELECT 2. Column or Expression List (SELECT Clause) (Type *in the SELECT clause.)
Select all rows?
SELECT (Omit the WHERE clause.)
Select only certain columns?
1. SELECT 2. Column or Expression List (SELECT Clause) (Type the list of columns in the SELECT clause.)
Select only certain rows?
1. SELECT 2. Conditions (Use a WHERE clause.)
Sort query results?
SELECT (Use an ORDER BY clause.)
Specify a foreign key?
Integrity (Use the ADD FOREIGN KEY clause in an ALTER TABLE command.)
Specify a primary key?
Integrity (Use the ADD PRIMARY KEY clause in an ALTER TABLE command.)
Specify a privilege?
GRANT
Specify integrity?
Integrity (Use an ADD CHECK, ADD PRIMARY KEY, and/or ADD FOREIGN KEY clause in an ALTER TABLE command.)
Specify legal values?
Integrity (Use an ADD CHECK clause in an ALTER TABLE command.)
Undo updates?
ROLLBACK
Update rows?
UPDATE
Use a calculated field?
1. SELECT 2. Column or Expression List (SELECT Clause) (Enter a calculation in the query.)
Use a compound condition?
1. SELECT 2. Conditions (Use simple conditions connected by AND, OR, or NOT in a WHERE clause.)
Use a compound condition in a query?
Conditions
FIGURE B-1
How Do I reference (continued)
How Do I Reference
292
How Do I
Review the Named Section(s) in Appendix A
Use a condition in a query?
1. SELECT 2. Conditions (Use a WHERE clause.)
Use a subquery?
Subqueries
Use a wildcard?
1. SELECT 2. Conditions (Use LIKE and a wildcard in a WHERE clause.)
Use an alias?
Aliases (Enter an alias after the name of each table in the FROM clause.)
Use set operations (union, intersection, difference)?
UNION, INTERSECT, and MINUS (Connect two SELECT commands with UNION, INTERSECT, or MINUS.)
FIGURE B-1
Appendix B
How Do I reference (continued)
APPENDIX
C
A N SWER S TO O DD - N UMBERED R E VIEW Q UESTIO N S CHAPTER 1—INTRODUCTION TO PREMIERE PRODUCTS, HENRY BOOKS, AND ALEXAMARA MARINA GROUP Due to the nature of the material in Chapter 1, there are no Review Questions.
CHAPTER 2—DATABASE DESIGN FUNDAMENTALS 1. 3.
5.
7.
An entity is a person, place, thing, or event. A relationship is an association between tables (entities). A one-to-many relationship between two tables is a relationship in which each row in the first table can be associated with many rows in the second table, but each row in the second table is associated with only one row in the first table. A relation is a two-dimensional table in which the entries in the table are singlevalued (each location in the table contains a single entry), each column has a distinct name (or attribute name), all values in a column match this name, the order of the rows and columns is immaterial, and each row contains unique values. For each table, you write the name of the table and then within parentheses list all of the columns in the table. Underline the primary keys. BRANCH (BRANCH_NUM, BRANCH_NAME, BRANCH_LOCATION, NUM_EMPLOYEES) PUBLISHER (PUBLISHER_CODE, PUBLISHER_NAME, CITY) AUTHOR (AUTHOR_NUM, AUTHOR_LAST, AUTHOR_FIRST) BOOK (BOOK_CODE, TITLE, PUBLISHER_CODE, TYPE, PRICE, PAPERBACK) WROTE (BOOK_CODE, AUTHOR_NUM, SEQUENCE) INVENTORY (BOOK_CODE, BRANCH_NUM, ON_HAND)
9.
A column (attribute), B, is functionally dependent on another column (or a collection of columns), A, if at any point in time a value for A determines a single value for B.
11.
Functional dependencies: DEPARTMENT_NUM DEPARTMENT_NAME ADVISOR_LAST_NAME, ADVISOR_FIRST_NAME, ADVISOR_NUM DEPARTMENT_NUM DESCRIPTION COURSE_CODE STUDENT_LAST_NAME, STUDENT_FIRST_NAME, STUDENT_NUM ADVISOR_NUM GRADE STUDENT_NUM, COURSE_CODE
294
Relations: DEPARTMENT (DEPARTMENT_NUM, DEPARTMENT_NAME) ADVISOR (ADVISOR_NUM, ADVISOR_LAST_NAME, ADVISOR_FIRST_NAME, DEPARTMENT_NUM) COURSE (COURSE_CODE, DESCRIPTION) STUDENT (STUDENT_NUM, STUDENT_LAST_NAME, STUDENT_FIRST_NAME, ADVISOR_NUM) STUDENT_COURSE (STUDENT_NUM, COURSE_CODE, GRADE)
Entity-relationship diagram: (Note: Your rectangles can be in different positions as long as they are connected by the same arrows.)
DEPARTMENT
ADVISOR
STUDENT
STUDENT_COURSE
COURSE
FIGURE C-1
13.
A table (relation) is in second normal form when it is in first normal form and no nonkey column is dependent on only a portion of the primary key. When a table is not in second normal form, the table contains redundancy, which leads to a variety of update anomalies. A change in a value can require not just one change, but several. There is the possibility of inconsistent data. Adding additional data to the database might not be possible without creating artificial values for part of the key. Finally, deletions of certain items can result in inadvertently deleting crucial information from the database.
15. STUDENT (STUDENT_NUM, STUDENT_LAST_NAME, STUDENT_FIRST_NAME, ADVISOR_NUM) ADVISOR (ADVISOR_NUM, ADVISOR_LAST_NAME, ADVISOR_FIRST_NAME) COURSE (COURSE_CODE, DESCRIPTION) STUDENT_COURSE (STUDENT_NUM, COURSE_CODE, GRADE) Appendix C
C H A P T E R 3 — C R E A T I N G TA B L E S 1. 3. 5.
7. 9. 11.
Use the CREATE TABLE command to create a table by typing the table name and then listing within a single set of parentheses the columns in the table. CHAR, VARCHAR, DATE, DECIMAL, INT, SMALLINT Answers will vary. Answers should mention that the difference between CHAR and VARCHAR is that CHAR is fixed length, while VARCHAR is variable length. This means that CHAR is always the same size and takes up the same amount of bytes, while VARCHAR varies. VARCHAR is a good choice when you are storing email addresses and comments that can vary in size. Use the INSERT command. Use the UPDATE command. Use the DESCRIBE command.
295
C H A P T E R 4 — S I N G L E - TA B L E Q U E R I E S 1.
3. 5.
7. 9. 11. 13. 15. 17.
The basic form of the SELECT command is SELECT-FROM-WHERE. Specify the columns to be listed after the word SELECT (or type * to select all columns), and then specify the table name that contains these columns after the word FROM. Optionally, you can include condition(s) after the word WHERE. You can form a compound condition by combining simple conditions and using the operators AND, OR, or NOT. Use arithmetic operators and write the computation in place of a column name. You can assign a name to the computation by following the computation with the word AS and then the desired name. In Oracle, the percent (%) wildcard represents any collection of characters. The underscore (_) wildcard represents any single character. Use an ORDER BY clause. To sort data in descending order, follow the sort key with the DESC operator. To avoid duplicates, precede the column name with the DISTINCT operator. Use a GROUP BY clause. Use the IS NULL operator in the WHERE clause.
C H A P T E R 5 — M U L T I P L E - TA B L E Q U E R I E S 1.
3. 5.
7.
Indicate in the SELECT clause all columns to display, list in the FROM clause all tables to join, and then include in the WHERE clause any conditions requiring values in matching columns to be equal. IN and EXISTS An alias is an alternate name for a table. To specify an alias in SQL, follow the name of the table with the name of the alias. You use the alias just like a table name throughout the SQL command. Use the UNION, INTERSECT, and MINUS operators to create a union, intersection, and difference of two tables. To perform any of these operations, the tables must be union compatible. Answers to Odd-Numbered Review Questions
9. 11. 296
13.
15.
When the ALL operator precedes a subquery, the condition is true only if it satisfies all values produced by the subquery. In an inner join, only matching rows from both tables are included. You can use the INNER JOIN clause to perform an inner join. In a right outer join, all rows from the table on the right will be included regardless of whether they match rows from the table on the left. Rows from the table on the left will be included only if they match. You can use the RIGHT JOIN clause to perform a right outer join. Answers will vary. Answers should note that an equi-join is similar to an inner join except that both matching columns appear in the results. A natural join is the same as the inner join discussed in Chapter 5. A cross join is the same as a Cartesian product.
CHAPTER 6—UPDATING DATA 1. 3. 5. 7.
9. 11. 13.
CREATE TABLE Use the INSERT command with a SELECT clause. DELETE In Oracle, use the ROLLBACK command. In SQL Server, use the ROLLBACK TRANSACTION command. Any updates made since the most recent COMMIT command (or COMMIT TRANSACTION command in SQL Server) are reversed. The clause is SET followed by the column name, followed by an equals sign (=) and the word NULL. In Oracle, use the ALTER TABLE command with a MODIFY clause. In SQL Server, use the ALTER TABLE command with an ALTER COLUMN clause. Use a make-table query to create a table from another table. The equivalent SQL commands to the Access make-table query are CREATE TABLE, SELECT, and INSERT.
CHAPTER 7—DATABASE ADMINISTRATION 1. 3. 5. 7. 9. 11.
13. 15.
Appendix C
A view contains data that is derived from existing base tables when users attempt to access the view. A defining query is the portion of the CREATE VIEW command that describes the data to include in a view. Views provide data independence, allow database access control, and simplify the database structure for users. DROP VIEW REVOKE Use the CREATE INDEX command to create an index. Use the CREATE UNIQUE INDEX command to create a unique index. A unique index allows only unique values in the column (or columns) on which the index is created. The DBMS Answers will vary. Answers should note that a data dictionary is a catalog that stores data about the entities, attributes, relationships, programs, and other
17.
19. 21.
objects in a database. Some items found in a data dictionary include synonyms for attributes, detailed descriptions of each table and attribute in the database, referential integrity constraints, and database schema definitions. The DBMS updates the system catalog automatically when users make change to the database, such as creating, altering, or dropping tables or creating or dropping indexes. Use the CHECK clause of the ALTER TABLE command. Use the ADD FOREIGN KEY clause of the ALTER TABLE command.
297
CHAPTER 8—SQL FUNCTIONS AND PROCEDURES 1.
3.
5.
7.
9. 11. 13. 15. 17.
19. 21.
Use the UPPER function to display letters in uppercase in Oracle and SQL Server. In Access, use the UCASE() function. Use the LOWER function to display letters in lowercase in Oracle and SQL Server. In Access, use the LCASE() function. To add months to a date, use the ADD_MONTHS function (Oracle), or the DATEADD() function (Access and SQL Server). To add days to a date, add the desired number of days to a date. To find the number of days between two dates, subtract the earlier date from the later date. In Oracle, separate the column names with two vertical lines (||) in the SELECT clause. In SQL Server, separate the column names with the + symbol. In Access, separate the column names with the & symbol. A stored procedure is a file that is stored on a server and contains commands that can be used repeatedly. Stored procedures eliminate the need for users to retype a query each time it is needed. In PL/SQL procedures, you declare variables first before any procedural code. Use the INTO clause to place the results of a SELECT statement in variables. When retrieving multiple rows with a SELECT statement, use a cursor. FETCH To use SQL commands in Access, create the command in a string variable. To run the command stored in the string variable, use the DoCmd.RunSQL command. To move to the next record in an Access recordset, use the MoveNext command. The INSERTED and DELETED tables are temporary system tables created by SQL Server. The INSERTED table contains the most recent (updated) values in a record and the DELETED table contains the previous (before update) value.
Answers to Odd-Numbered Review Questions
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INDEX @repnum, 257–258
ADD PRIMARY KEY clause, 222 adding, 207
A
rows, 207 additions, 44
Access (Microsoft), 187
to tables, 44–45, 48
ALTER TABLE command, 187
ADD_MONTHS function, 237
and PL/SQL programs, 234
aggregate functions, 114
changing column names in, 200
Alexamara Marina Group (example), 169
character functions in, 236 concatenating columns in, 241 creating indexes in, 217 creating views in, 197 data types in, 172, 279
exercises, 169 Alexamara Marina Group database (example), 231–232 exercises, 59–60, 94, 134, 192–193, 271–272
Documenter, 281
introduction to, 2, 15, 19
Documenter tool, 86, 186, 218
sample data, 15, 19
parameter queries in, 242 query results display in, 99 rollbacks in, 180–181 running SQL commands, 68 saving SQL commands, 82 SELECT command, 77 specifying foreign keys in, 223 specifying primary keys in, 222 using SQL in, 259, 263 validation rules in, 224 wildcards in, 111, 276 working with dates in, 238, 240 access control, 209 security and, 209, 212 ADD clause, 183 ADD FOREIGN KEY clause, 222
aliases, 273 for self-joins, 147 using, 146 ALL operator, 157, 159, 277 ALTER COLUMN clause, 187 ALTER TABLE command, 182, 188, 273–274 ADD PRIMARY KEY clause, 222 CHECK clause, 224 integrity constraints with, 221–222 specifying integrity with, 283–284 AND condition, 103 AND operator, 103, 276 ANY operator, 157, 159, 277 arguments, 235 arithmetic operators, 107
300
AS clause, 108, 117
columns, 183
asterisk (*), 100, 111
adding, 183
attributes, 26
changing values to null, 181–182
database, 26, 28
computed, 107, 110, 274
identifying, 34
concatenating, 240, 242
Autocommit, 177, 181
decreasing length of, 187
AVG function, 116, 275
determinant, 48–49 in databases, 28
B
listing, 88, 281 nokey, 45 renaming, when creating views, 199–200
base tables, 196 joins of, updating views of, 205, 208 BETWEEN operator, 106–107, 276 Boyce-Codd normal form (BCNF), 49 breadcrumbs, 66
retrieving, 98, 100 retrieving single, 242, 244, 256 selecting, 274 using self-joins on primary key, 148, 150 commands, 95–96 reversing, 284–285
C calling, 244 stored procedures, 244 candidate keys, 33, 48–49 Cartesian products, 163–164 categories, 2 relationships between, 2 changing, 247 rows, 247 CHAR data type, 71 character functions, 235–236 CHECK clause, 224 child table, 223 client, 242
comparison operators, 101 compound conditions, 103, 106, 276 in updates, 175 computations, 107, 110 computed columns, 107, 110, 274 concatenation, 30 of columns, 240, 242 conditions, 275, 277 ALL, 277 AND, 103 ANY, 277 BETWEEN, 276 compound, 103, 106, 276 EXISTS, 277
client/server systems, 242
IN, 276
CLOSE command, 249, 252, 257–258
LIKE, 276
column names, 26
NOT, 105–106
conventions for, 26 qualifying, 29, 136–137
Index
COMMIT command, 171, 177–178, 275
nulls in, 128
301
OR, 104
inconsistent, 44, 48
simple, 101–102, 104, 275
inserting, with Visual Basic, 262
correlated subqueries, 142
sorting, 112, 114
COUNT function, 114–115, 117, 275
updating, using views, 203, 208
CREATE INDEX command, 215, 217, 277 CREATE PROCEDURE command,
updating, with Visual Basic, 261–262 data dictionary, 218 data types, 66, 279
243–244, 256 CREATE TABLE command, 66, 69, 72, 82,
assigning to variables, 244 using, 71
86, 188, 278–279 integrity constraints with, 221–222
database administration, 196
table structure and, 87–88
creating and using views, 196, 203
with errors, 69–70
indexes and, 212, 217
CREATE UNIQUE INDEX command, 217
integrity constraints and, 221, 225
CREATE VIEW command, 196, 201
introduction to, 195–196
creating, 215
security and, 209, 212
indexes, 215, 217
system catalog and, 218, 221
stored procedures, 242, 244
updating data using views, 203, 208
CURRENCY data type, 68, 172, 279
database administrator, 195
cursors, 255
database design, 52
advantages of, 255
diagrams for, 52, 55
closing, 252
introduction to, 23
complete procedure using, 252–253
method for, 34–35
complex, 254–255, 258–259
process example, 36, 41
fetching rows from, 251–252
relations and, 26, 29 requirements, 35–36
opening, 250 using, 249–250, 257–258
databases, 24 concepts of, 24, 29
D data, 173
defined, 2 Datasheet view, 68 DATE data type, 71
changing, in table, 173, 175
DATE() function, 240
changing, with stored procedure,
DATEADD() function, 238
247, 256 customizing display of, 203
dates, 237 working with, 237, 240
deleting, with procedure, 248, 257
DBA_TABLES, 218
deleting, with Visual Basic, 259–260
DBA_TAB_COLUMNS, 218
grouping, 123, 128
DBA_VIEWS, 218 DBMS_OUTPUT, 244 Index
302
DEALLOCATE command, 257–258 Debug.Print command, 260
error handling, 245 with stored procedures, 245–246
DECIMAL data type, 68, 71, 172, 279
EXCEPTION clause, 245–246
DECLARE statement, 257–258
EXEC command, 256
defining query, 196, 199
EXISTS operator, 277
DELETE command, 171, 178, 181, 208,
and joins, 141–142
248–249, 259–260, 280 deleting, 259
F
data with Visual Basic, 259–260 data, with procedure, 248, 257
257–258
indexes, 281 rows, 178, 181, 280
fields, 28 in databases, 28
tables, 281 views, 282
first normal form (1NF), 41, 43
deletions, 45
FLOOR function, 237
from database, 45, 48 DESC operator, 113
foreign keys, 222–223 FROM clause, 98
DESCRIBE command, 87, 187–188, 281
and joins, 151
determinant, 48–49
full outer joins, 161
diagrams, 52
functional dependence, 29, 31, 34
entity-relationship (E-R), 52, 55
functions, 275
DISTINCT operator, 117, 119, 204, 274
aggregate, 114
DoCmd.RunSQL command, 259
AS clause with, 117
Documenter, 281
AVG function, 116
Documenter tool, 86, 186, 218
character, 235–236
DROP INDEX command, 217, 281
COUNT function, 114–115, 117
DROP TABLE command, 70, 189, 281
introduction to, 233–234
DROP VIEW command, 208–209, 282
MAX function, 116 MIN function, 116
dropping, 217
number, 236–237
indexes, 217
SUM function, 115–116
views, 208–209
using, 114, 119, 235, 240
E entities, 26, 34 entity-relationship (E-R) diagrams, 52, 55
Index
FETCH command, 249, 251–252, 254,
G GETDATE() function, 240 GRANT command, 209, 212, 282
303
GROUP BY clause, 123, 125, 145
INT data type, 71
grouping data, 123
integrity, 283–284
in databases, 123, 128
integrity constraints, 221, 225 integrity support, 221–222
H
INTERSECT operator, 152, 155–156, 286–287
HAVING clause, 125, 128
IS NOT NULL operator, 128
Henry Books database (example), 229, 231
IS NULL operator, 128
exercises, 59, 91, 133–134, 168–169, 192, 270–271
J
introduction to, 2, 8, 14 sample data, 8, 13
joining, 143 multiple tables, 143–144, 150, 152
I
two tables, 136, 142 joins, 160
Immediate window, 260–261
inner, 160–161
IN clause, 111–112
outer, 161, 163
IN operator, 111–112, 276
self-joins, 147, 149
and joins, 139, 141
updating views involving, 205, 208
using, 140–141 inconsistent data, 44, 48
K
indexes, 215 creating, 215, 217, 277
key, 112–113
deleting, 281 described, 212, 215
L
dropping, 217 unique, 217
LCASE function, 236
inner joins, 160–161
left outer joins, 161
INSERT command, 72, 75, 82, 86, 171, 188,
LIKE conditions, 276
203–204, 207, 262 adding rows using, 176–177
LIKE operator, 110–111 LOWER function, 236
in CUSTOMER table, 173 INSERT INTO (query) command, 282
M
INSERT INTO (values) command, 283 inserting, 283 inserting, 262 data, with Visual Basic, 262
major sort key, 113 MAX function, 116, 275 MIN function, 116, 275 Index
304
minor sort key, 113
O
MINUS operator, 152, 156–157, 286–287 MODIFY clause, 186, 188
one-to-many relationships, 26
multiple-table queries, 135, 169
OPEN command, 249–250, 257–258 operations, 163
N names, 108 assigning, to computed columns, 108
product, 163–164 set, 152, 157 special, 160, 164 optimization, 144
naming conventions, 26
OR condition, 104
nested subqueries, 142, 144
OR operator, 103, 276
nesting, 120
Oracle, 163
queries, 120, 123 nokey column, 45 nonprocedural languages, 234, 272 normal forms, 41 normalization, 41, 52 Boyce-Codd normal form (BCNF), 49
outer joins in, 163 system catalog and, 218 Oracle Database Express Edition, 99 query results display in, 99 starting, 62, 64 order, 6
first normal form (1NF), 41, 43
ORDER BY clause, 112, 114, 136, 201
goal of, 41
ORDERS table, 6–7
second normal form (2NF), 43, 47
outer joins, 161, 163
third normal form (3NF), 47, 52 NOT condition, 105–106
P
NOT NULL clause, 72 NOT operator, 103, 276
parameter queries, 242
null data value, 72
parent table, 223
null values, 128
percent (%) wildcard, 276
in conditions, 128
percent sign (%), 110
in SUM, AVG, MAX, MIN functions, 117
PL/SQL, 234
nulls, 183 changing values to, 181–182, 187–188
cursors in, 249, 255
inserting rows with, 75
error handling in, 245–246
using, 72
stored procedures in, 245, 255
NUMBER data type, 68
PL/SQL commands, 243
number functions, 236–237
Premier Products database (example), 228
numbers, 106 in queries, 106 Index
creating stored procedures, 242, 244
exercises, 228–229 integrity constraints in, 221–222
305
Premiere Products database (example), 36
saving as stored procedures, 242, 244
design process for, 36, 41
subqueries, 121, 123
E-R diagram of, 52, 55
using aliases in, 146
exercises, 58–59, 90–91, 132–133,
using SELECT command, 98, 112
167–168, 191–192, 269–270
question mark (?), 111
introduction to, 2, 8 sample data, 3, 6, 25
R
sample order, 2–3 table structure, 6–7 primary key columns, 148 using self-joins on, 148, 150
records, 28 in databases, 28 redundancy, 44
primary keys, 222
relational databases, 24
adding, 222
concepts of, 24, 29
in databases, 31, 34
functional dependence in, 29, 31, 34
normal forms and, 45, 47
introduction to, 24, 26
shorthand representation of, 33
normalization of, 41, 52
with joins, 208
shorthand for, 28–29
primary sort key, 113 privileges, 282
relations, 27 in databases, 27–28
granting to users, 282 revoking, 284
unnormalized, 41 relationships, 26
user, 209, 212
in databases, 26, 28
procedural code, 244
one-to-many, 26
procedural languages, 234
repeating groups, 27–28
procedures, 272
review questions, 227
products, 163–164
database administration, 227–228 multiple-table queries, 166–167
Q queries, 144
single-table queries, 131–132 SQL, 89–90 updating data, 190–191
comprehensive example, 144–145
REVOKE command, 211–212, 284
constructing simple, 98, 112
right outer joins, 161
defined, 98
ROLLBACK command, 171, 177–178,
defining, 196 multiple-table, 135, 169 nesting, 120, 123 parameter, 242
180–181, 284–285 rollbacks, 180 executing, 180–181 ROUND function, 236–237 Index
306
row-and-column subset views, 203 updating, 203, 205
introduction to, 97 retrieving multiple rows using, 249, 262–263
rows, 176 adding, 176–177, 207
sorting data with, 112–113
changing, with update procedure, 247
WHERE clause in, 100, 103
counting, 114–115
with compound conditions, 103, 106
cursors with, 249, 255
with computed columns, 107, 110
deleting, 178, 181, 280
with GROUP BY clause, 123, 125
deleting, with procedure, 248
with IN clause, 111–112
grouping, 123, 128
with LIKE operator, 110–111 self-joins, 148
in databases, 28 inserting, 72, 75, 282–283
incorrect, 148
restricting, in join, 138–140
on primary key column, 148, 150 using, 147–148
retrieving, 98, 100 retrieving single, 242, 244, 256
SEQUEL, 61
selecting multiple, 249, 255, 262–263
server, 242
updating, 178, 181, 274
set operations, 152, 157
RTRIM function, 240–241
SET SERVEROUTPUT ON command, 244 simple conditions, 101–102, 104, 275
S
slash (/), 244 SMALLINT data type, 71
script files, 242–243 script repository, 80
as primary key, 33
scripts, 63–64, 80, 82
sort key, 112–113
second normal form (2NF), 43, 47
sorting, 112, 114
secondary sort key, 113
sp_columns stored procedure, 88
security, 209
sp_tables procedure, 218
access control and, 209, 212 provided by views, 203 SELECT clause, 98 ALL operator with, 157, 159 and joins, 145, 151–152 ANY operator with, 158 SELECT command, 75, 78, 285 counting rows using, 114–115 described, 98 for column and row retrieval, 98, 100
Index
Social Security numbers, 33
specific functions, 272 SQL, 221 integrity constraints in, 221, 225 SQL (Structured Query Language), 234 extensions, 234 functions, 235, 240 introduction to, 61–62 scripts, 63–64 using in Access, 259, 263 using in programming environment, 234
307
SQL commands, 95–96
in Visual Basic, 259, 263
comparison operators in, 101
triggers, 264, 267
correcting errors in, 69–70
update procedures, 247–248, 256
creating and running, 62, 66
using cursors in, 249, 255
embedding, in another language, 234
subqueries, 121, 123, 286
entering, 65–66
ALL and ANY operators with, 157, 159
in Access, 68
correlated, 142
saving, 80, 82
nested, 142, 144
storing in string variable, 259–260
SUM (BALANCE) function, 235
triggers with, 259, 263
SUM function, 115–116, 275
SQL Scripts option, 63–64
SYSCOLUMNS, 218
SQL Server, 187
SYSDATE function, 239
ALTER COLUMN clause, 187
SYSTABLES, 218
COMMIT and ROLLBACK
system catalog, 218, 221
commands, 178
SYSVIEWS, 218
concatenating columns in, 242 listing columns in, 186
T
nulls in, 182 ordering query results in, 201 query results display in, 99 stored procedures, 218 triggers in, 266 using T-SQL in, 256, 259 working with dates in, 238, 240 SQL Server (Microsoft), 69 scripts, 82 SELECT command, 78 sp_columns, 88 SQL Server Management Studio, 69 statistics, 208 updating views involving, 208 stored procedures, 218 calling, 244 described, 242 error handling, 245–246 for selecting multiple rows, 249, 255 illustration of, 242, 244
T-SQL, 272 Table Design view, 68 tables, 72 adding rows, 72, 75 adding rows to, 176–177 aliases for, 146 base, 196 changing data in existing, 173, 175 changing structure of, 182, 188 changing values to null in, 181–182 correcting errors in, 78, 80 creating, 66, 69, 82, 86, 278–279 creating new from existing, 172–173 data types, 71 deleting, 281 deleting rows from, 178, 181, 248 describing, 87–88 difference of, 152, 156–157 dropping, 70, 189
in T-SQL, 256, 259 Index
308
foreign keys for, 222–223
unnormalized relations, 41
intersection of, 152, 155–156
update anomalies, 44, 48
joining multiple, 143–144
UPDATE command, 78, 80, 171, 173, 175,
joining several, 150, 152
179, 181, 183–184, 188, 208, 247, 249,
joining two, 136, 142
260–261, 274
naming conventions, 26
update procedures, 247–248, 256
normalization of, 41, 52
updates, 177
primary keys of, 31, 34
committing, 177–178
product of two, 163–164
reversing, 284–285
querying multiple, 135, 169
roll backs, 177–178
redundancy in, 44
to tables, 44, 48
relations in, 26, 29
updating, 203
repeating groups in, 27
data, using views, 203, 208
self-joins, 147, 149
data, with Visual Basic, 260–261
system catalog and, 218, 221
rows, 178, 181
union compatible, 153
table data, 173, 175
union of, 152–153, 155
table structure, 182, 188
updating rows in, 178, 181 viewing data, 75, 78 third normal form (3NF), 47, 52 Transact-SQL (T-SQL), 234 cursors in, 257, 259
UPPER function, 235–236 user access, 209, 212 user privileges, 282 granting, 282 revoking, 284
retrieving single row and column, 256 stored procedures in, 256, 259
V
using in SQL Server, 256, 259 transactions, 178
VALUES command, 75
triggers, 264, 267
VARCHAR data type, 71
tuples, 28
variable names, 244 in PL/SQL, 244
U
variables, 244 assigning data type to, 244
UCASE function, 236 underscore (_), 111
Index
views, 202 benefits of, 202–203
underscore (_) wildcard, 276
creating, 279
UNION operator, 152–153, 155, 286–287
creating and using, 196, 203
union-compatible tables, 153
deleting, 282
unique indexes, 217
dropping, 208–209
309
involving joins, 205, 208
W
involving statistics, 208 row-and-column subset views, 203, 205 updating data using, 203, 208 Visual Basic, 259
WHERE clause, 136 and joins, 136, 138–139, 150–151 described, 98
deleting data with, 259–260
using, 100, 103
finding multiple rows with, 262–263
vs. HAVING clause, 125, 128
inserting data with, 262
wildcards, 110–111, 276
running code in, 260–261
WITH GRANT OPTION clause, 211–212
updating data with, 260–261
Index