chapter 4 constraints
DESCRIPTION
Chapter 4 Constraints. Jason C. H. Chen , Ph.D. Professor of MIS School of Business Gonzaga University Spokane, WA 99258 USA [email protected]. What will we study today?. Data. Integrity. How to achieve it?. Referential Integrity. …. …. Objectives. - PowerPoint PPT PresentationTRANSCRIPT
Dr. Chen, Oracle Database System (Oracle) 1
Oracle_ch3 HW (#7 & 8)
SQL> --7.SQL> CREATE TABLE book_pricing (id, cost, retail, category) 2 AS (SELECT isbn, cost, retail, category 3 FROM books);
Table created.SQL> --7b.(version 2)SQL> SQL> DROP TABLE book_pricing CASCADE CONSTRAINTS;
Table dropped.
SQL> SQL> CREATE TABLE book_pricing 2 AS (SELECT isbn as id, cost, retail, category 3 FROM books);
Table created.
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SQL> --8.SQL> ALTER TABLE book_pricing 2 SET UNUSED (category);Table altered.
SQL> -- use one of the following commands to verify the resultSQL> SQL> --8aSQL> DESCRIBE book_pricing; Name Null? Type ----------------------------------------- -------- ---------------------------- ID VARCHAR2(10) COST NUMBER(5,2) RETAIL NUMBER(5,2)SQL> --8bSQL> SELECT * FROM book_pricing;
ID COST RETAIL ---------- ---------- ---------- 1059831198 18.75 30.95 0401140733 14.2 22 4981341710 37.8 59.95 8843172113 31.4 55.95 3437212490 12.5 19.95 3957136468 47.25 75.95 1915762492 21.8 25 9959789321 37.9 54.5 2491748320 48 89.95 0299282519 19 28.75 8117949391 5.32 8.95
ID COST RETAIL ---------- ---------- ---------- 0132149871 17.85 29.95 9247381001 15.4 31.95 2147428890 21.85 39.95
14 rows selected.
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What will we study today?
IntegrityData
… … ReferentialIntegrity
(Constraints)
How to achieve it?
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Chapter 4Constraints
Jason C. H. Chen, Ph.D.Professor of MIS
School of BusinessGonzaga University
Spokane, WA 99258 [email protected]
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Objectives
• Explain the purpose of constraints in a table• Distinguish among PRIMARY KEY, FOREIGN
KEY, UNIQUE, CHECK, and NOT NULL constraints and the appropriate use for each constraint
• Understand how constraints can be created when creating a table or modifying an existing table
• Distinguish between creating constraints at the column level and table level
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Objectives (continued)
• Create PRIMARY KEY constraints for a single column and a composite primary key (cpk)
• Create a FOREIGN KEY constraint• Create a UNIQUE constraint• Create a CHECK constraint• Create a NOT NULL constraint using the ALTER
TABLE…MODIFY command• Include constraints during table creation• Use DISABLE and ENABLE commands• Use the DROP command
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Referential Integrity• Move to JLDB_Referential_Integrity.pptx
file
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customers
orders
pk
fkpk
customer# LastName … Referred Region1001 MORALES NULL SE… …
1005 GIRARD NULL NW1020 FALAH NULL NE
Order# customer# … ShipZip ShipCost1000 1005 98114 2.00… … …
1003 1001 32328 4.001012 1007 49002 6.00
Can we “create” orders#1 if customers#5 is not created? Why?
Can we “delete” customers#5 if orders#1 is still in the database? Why?
Customers#5
orders#1
Referential Integrity
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Refresh the Database
• 1. Create a new folder on c:\ as follows:c:\oradata\chapter4
• 2. Go to Blackboard and download data files from Oracle chapter4 and save under c:\oradata\chapter4\
• 3. Run the following script file– Start c:\oradata\chapter4\JLDB_Build_4.sql
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Type the following SQL commands
-- chapter 4, Figure 4-5; p. 108INSERT INTO customers (customer#, lastname,
firstname, region)VALUES (1020, 'PADDY', 'JACK', 'NE');--extra INSERT commandINSERT INTO orders (order#, customer#, orderdate)VALUES (1021, 1021, '06-APR-09');-- DELETE commandDELETE FROM customersWHERE customer# = 1005;
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Constraint Checked with Data Input
Figure 4-5 Insert a row to test the constraintWhat cause the problem?
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customers
RULES: 1. You can’t add a record to TABLE- (or the table with fk, e.g., orders)
unless there is a corresponding record in TABLE-1 (or the table with pk).
2. You can’t delete a record in TABLE-1 (or the table with pk, e.g., customers) if there is a record in TABLE- (or the table with fk).
Order of entering data into the database: customers orders Order of deleting data from the database: orders customers
orders
pk
fkpk
customer# LastName … Referred Region1001 MORALES NULL SE… …
1005 GIRARD NULL NW1020 FALAH NULL NE
Order# customer# … ShipZip ShipCost1000 1005 98114 2.00… … …
1003 1001 32328 4.001012 1007 49002 6.00
Using the FOREIGN KEY ConstraintReferential Integrity
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Review from Last Class
What are the three rules of naming table and field?
What are the three (total of four) required information that should be described for each field?
1. Name 2. Type3. Size4. Constraint
L
(saved in “UPPER” case in the D.B.)
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What is a Constraint?
• A rule used to enforce business rules, practices, and policies
• A rule used to ensure accuracy and integrity of data
• A mechanism used to protect – the relationship between data within an Oracle
table, or– the correspondence between data in two different
tables.– For example, the state entered must be one of the 50
states in the U.S.
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Types of Constraints
• Integrity constraints: define primary and foreign keys
• Value constraints: define specific data values or data ranges that must be inserted into columns and whether values must be unique or not NULL
• Table constraint: restricts the data value with respect to all other values in the table
• Field (column) constraint: limits the value that can be placed in a specific field, irrespective of values that exist in other table records
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I. Naming conventions for constraints
<tablename>_<fieldname>_<constraint id>Where <constraint id> is:
• pk PRIMARY KEY• fk REFERENCES <tablename> (pk)• ck CHECK <condition to be checked>
(note that cc stands for CHECK CONDITION)• nn NOT NULL• uk UNIQUE
Inte
grai
tyco
nstr
aint
Valu
eco
nstr
aint
e.g., s_id NUMBER (6) CONSTRAINT student_s_id_pk PRIMARY KEY;
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Integrity Constraints
• Define primary key fields
•Specify foreign keys and their corresponding table and column references
•Specify composite keys
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Creating a Table
CREATE TABLE tablename(fieldname1 data_type (size)
[CONSTRAINT constraint_name constraint_type], fieldname2 data_type (size),
…[CONSTRAINT constraint_name constraint_type,]
…);
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Creating Constraints
• When– During table creation– After table creation, by modifying the existing
table• How
– Column level approach– Table level approach
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Creating Constraints at the Column Level
• If a constraint is being created at the column level, the constraint applies to the column specified
Creating Constraints at the Table Level• Approach can be used to create any constraint type after all
table field definitions are completed except NOT NULL• Required if constraint is based on multiple columns
Figure 4-1 Syntax for creating a column-level constraint
Figure 4-2 Syntax for creating a table-level constraint
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Customer# LastName FirstName Address City State Zip Referred Region EmailNUMBER(4)
VARCHAR2(10) VARCHAR2(10)
VARCHAR2(20)
VARCHAR2(12) VARCHAR2(2)
VARCHAR2(5)
NUMBER(4)
CHAR(2) VARCHAR2(30)
Order# Customer# OrderDate ShipDate ShipStreet ShipCity ShipState ShipZip ShipCostNUMBER(4) NUMBER(4) DATE DATE VARCHAR2(18) VARCHAR2(15) VARCHAR2(2) NUMBER(4) NUMBER(4,2)
CREATE TABLE Customers(Customer# NUMBER(4),LastName VARCHAR2(10) NOT NULL,FirstName VARCHAR2(10) NOT NULL,Address VARCHAR2(20),City VARCHAR2(12),State VARCHAR2(2),Zip VARCHAR2(5),Referred NUMBER(4),Region CHAR(2),Email VARCHAR2(30), CONSTRAINT customers_customer#_pk PRIMARY KEY(customer#), CONSTRAINT customers_region_ck CHECK (region IN ('N', 'NW', 'NE', 'S', 'SE', 'SW', 'W', 'E')) );
CUSTOMERSpk
ORDERSfkpk
Optional (variable name)
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Qs
• Q: How to display (describe) the table structure you just created of “customers” table?
• A: _______________________• Q: Any constraint(s) displayed?• A: ____• Q: How to display them?• A:_
DESCRIBE customers;
SELECT constraint_name FROM user_constraints WHERE table_name=‘CUSTOMERS’;
NO! case is sensitive(UPPER case) Why?
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Customer# LastName FirstName Address City State Zip Referred Region EmailNUMBER(4)
VARCHAR2(10) VARCHAR2(10)
VARCHAR2(20)
VARCHAR2(12) VARCHAR2(2)
VARCHAR2(5)
NUMBER(4)
CHAR(2) VARCHAR2(30)
Order# Customer# OrderDate ShipDate ShipStreet ShipCity ShipState ShipZip ShipCostNUMBER(4) NUMBER(4) DATE DATE VARCHAR2(18) VARCHAR2(15) VARCHAR2(2) NUMBER(4) NUMBER(4,2)
CREATE TABLE Orders (Order# NUMBER(4), Customer# NUMBER(4), OrderDate DATE NOT NULL, ShipDate DATE, ShipStreet VARCHAR2(18), ShipCity VARCHAR2(15), ShipState VARCHAR2(2), ShipZip VARCHAR2(5),ShipCost NUMBER(4,2), CONSTRAINT orders_order#_pk PRIMARY KEY(order#), CONSTRAINT orders_customer#_fk FOREIGN KEY (customer#) REFERENCES customers(customer#));
CUSTOMERSpk
ORDERSfkpk
Optional (variable name)
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Order# Customer# OrderDate ShipDate ShipStreet ShipCity ShipState ShipZip ShipCostNUMBER(4) NUMBER(4) DATE DATE VARCHAR2(18) VARCHAR2(15) VARCHAR2(2) NUMBER(4) NUMBER(4,2)
ORDERS
Order# Item# ISBN Quantity PaidEachNUMBER(4) NUMBER(2) VARCHAR2(10) NUMBER(3) NUMBER(5,2)
ORDERITEMS
CREATE TABLE ORDERITEMS ( Order# NUMBER(4), Item# NUMBER(2), ISBN VARCHAR2(10), Quantity NUMBER(3) NOT NULL, PaidEach NUMBER(5,2) NOT NULL, CONSTRAINT orderitems_order#item#_pk PRIMARY KEY (order#, item#), CONSTRAINT orderitems_order#_fk FOREIGN KEY (order#) REFERENCES orders (order#) , CONSTRAINT orderitems_isbn_fk FOREIGN KEY (isbn) REFERENCES books (isbn) , CONSTRAINT oderitems_quantity_ck CHECK (quantity > 0) );
ISBN Title PubDate PubID Cost Retail Discount CategoryVARCHAR2(10) VARCHAR2(30) DATE NUMBER(2) NUMBER(5,2) NUMBER(5,2) NUMBER(4,2) VARCHAR2(12)
BOOKSpk
fkfk
pk
Optional (variable name)
cpk
How to define ‘composite” key?
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Your Job
• You need to study and understand all CREATE TABLE SQL commands in JLDB_Build_4.sql
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Enforcement of Constraints
• All constraints are enforced at the table level
• If a data value violates a constraint, the entire row is rejected
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Constraint Types
Table 4-1 Constraint types
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Your Turn …
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Syntax:CONSTRAINT constraint_name PRIMARY KEY
Primary Key Constraints
SQL> CREATE TABLE students2 (s_id NUMBER(6) CONSTRAINT students_s_id_pk PRIMARY KEY,3 s_name VARCHAR2(30),4 s_class CHAR(2),5 s_dob DATE);
Create a table with the following information:1. Name of the table: students2. Fields: s_id number with 6 digits and is a primary key, s_name
character with 30 chars, s_class with 2 chars, s_dob with DATE
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at the Column-Level
Primary Key Constraints (cont.)
SQL> CREATE TABLE students2 (s_id NUMBER(6), 3 s_name VARCHAR2(30),4 s_class CHAR(2),5 s_dob DATE,6 CONSTRAINT students_s_id_pk PRIMARY KEY (s_id));
SQL> CREATE TABLE students2 (s_id NUMBER(6) CONSTRAINT students_s_id_pk PRIMARY KEY,3 s_name VARCHAR2(30),4 s_class CHAR(2),5 s_dob DATE);
Practice: Type in one of the command.
at the Table-Level
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Your Job
• Read the rest of powerpoint slide and practice all examples
• Skip to another practice example with THREE new tables
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Adding Constraints to Existing Tables
• Constraints are added to an existing table with the ALTER TABLE command
• Add a NOT NULL constraint using MODIFY clause
• All other constraints are added using ADD clause
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Using the PRIMARY KEY Constraint
• Ensures that columns do not contain duplicate or NULL values
• Only one per table is allowed
Figure 4-3 Syntax of the ALTER TABLE command to add a PRIMARY KEY constraint
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Drop Contraint
ALTER TABLE orderitemsDROP CONSTRAINT
orderitems_order#item#_pk PRIMARY KEY (order#, item#);
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Deletion of Foreign Key Values
• You cannot delete a value in a parent table referenced by a row in a child table
• Use ON DELETE CASCADE keywords when creating FOREIGN KEY constraint – it automatically deletes a parent row when the row in a child table is deleted
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Practice …
• Let’s try to create additional tables– JustLee Books would like to create some new
tables to store office equipment inventory data.
DeptIDDnameFax
DEPTEquipIDEdescPurchdateRatingDeptIDEtypeID
EQUIP
EtypeIDEtypename
ETYPES
Figure 4-26 E-R model for equipment tables
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Including Constraints during Table Creation
DeptIDDnameFax
DEPTEquipIDEdescPurchdateRatingDeptIDEtypeID
EQUIP
EtypeIDEtypename
ETYPES
• Each department name must be unique.• Each department must be assigned a name .
• Each equipment type name must be unique• Each equipment type must be assigned a name.• Each equipment item must be assigned a valid department.• If an equipment item is assigned a type, it must be a valid type.• Valid rating values for equipment are A, B, and C.
uniqueNOT NULL
uniqueNOT NULL
ck
fk fk
pkpk pk
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-- chapter 4, Figure 4-27; p. 122CREATE TABLE dept(deptid NUMBER(2), dname VARCHAR2(20) NOT NULL, fax VARCHAR2(12), CONSTRAINT dept_deptid_pk PRIMARY KEY (deptid), CONSTRAINT dept_dname_uk UNIQUE (dname) );
DEPT table creation
• Each department name must be unique.• Each department must be assigned a name .
uniqueNOT NULL
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-- chapter 4, Figure 4-27; p. 122CREATE TABLE dept(deptid NUMBER(2), dname VARCHAR2(20) NOT NULL, fax VARCHAR2(12), CONSTRAINT dept_deptid_pk PRIMARY KEY (deptid), CONSTRAINT dept_dname_uk UNIQUE (dname) );
-- chapter 4, Figure 4-30; p. 124CREATE TABLE dept(deptid NUMBER(2) CONSTRAINT dept_deptid_pk PRIMARY KEY, dname VARCHAR2(20) NOT NULL CONSTRAINT dept_dname_uk UNIQUE, fax VARCHAR2(12));
Constraints are defined at the column-level
What is the main difference on the following “CREATE TABLE” statements?
Constraints are defined at the table-level
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-- chapter 4, Figure 4-28; p. 123CREATE TABLE etypes(etypeid NUMBER(2), etypename VARCHAR2(20) NOT NULL, CONSTRAINT etypes_etypeid_pk PRIMARY KEY (etypeid), CONSTRAINT etypes_etypename_uk UNIQUE (etypename) );
-- chapter 4, Figure 4-29; p. 123CREATE TABLE equip(equipid NUMBER(3), edesc VARCHAR2(30), purchdate DATE, rating CHAR(1), deptid NUMBER(2) NOT NULL, etypeid NUMBER(2), CONSTRAINT equip_equipid_pk PRIMARY KEY (equipid), CONSTRAINT equip_deptid_fk FOREIGN KEY (deptid) REFERENCES dept (dept_id), CONSTRAINT equip_etypeid_fk FOREIGN KEY (etypeid) REFERENCES etypes (etypeid), CONSTRAINT equip_rating_ck CHECK (rating IN ('A', 'B', 'C', 'D')) );
• Each equipment item must be assigned a valid department.• If an equipment item is assigned a type, it must be a valid
type.• Valid rating values for equipment are A, B, and C.
• Each equipment type name must be unique• Each equipment type must be assigned a name.
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• HW!!• Practice all the examples in the text.• A Script file is available on the Bb (file name:
Ch4Queries.sql)• After completing all examples, do the HW (hint: see
the tables below for the final schema and the sample output on the Bb under “Assignments”
rep_id last first comm base_salaryNUMBER(5) VARCHAR2(15) VARCHAR2(10) CHAR(1) NUMBER(7,2)
store_id name contact rep_idNUMBER(8) VARCHAR2(30) VARCHAR2(30) NUMBER(5)
store_id name quarter rep_idNUMBER(8) VARCHAR2(5) CHAR(3) NUMBER(5)
pk store_reps
pk fkbook_stores
cpk, fk cpk cpk, fkrep_contracts
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Homework - Hands-On Assignments
Read and Practice all examples on Chapters 4• 1. Run the script files (in the folder \oradata\
chapter4\): JLDB_Build_4.sql• 2. Read Oracle assignment and create a script file
Oracle_ch4_Lname_Fname.sql for questions (#1 to #8; p.133) on “Hands-on Assignments”. .
• 3. Execute and test one problem at a time and make sure they are all running successfully.
• 4. When you done, spool the script files (see next slide for spooling instructions) and UPLOAD both *.sql and spooled file Oracle_ch4_Spool_Lname_Fname.txt to Bb by the midnight before the next class.
• Turn in a hardcopy of spooled file (*.txt ONLY) to me in the class.
Upload the SQL and spooled files (*.sql and *.txt) to the Bb (under “Assignments & Projects”) by the deadline
[correction on p.138: Name Datatype should be VARCHAR2(5)]
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How to Spool your Script and Output FilesAfter you tested the script file of Oracle_ch4_Lname_Fname.sql successfully,
follow the instructions below to spool both script and output files:Step 0. Run the following script file from SQL*Plus (since you have created
JLDB tables)– Start c:\oradata\chapter4\JLDB_Build_4.sql
• 1. type the following on SQL>– Spool c:\oradata\Oracle_ch4_Spool_Lname_Fname.txt (make sure your name is entered)
• 2. open Oracle_ch4_Lname_Fname.sql that you already tested• 3. copy and paste all the SQL commands (including all comments) to the
SQL*PLUS • 4. type Spool Off on the SQL>The output should contain your personal information, all SQL commands and
their solution on the .txt file and saved in C: drive (oradata\ folder)
Upload the SQL and spooled files (*.sql and *.txt) to the Bb (under “Assignments & Projects”) by the deadline
Dr. Chen, Oracle Database System (Oracle) 44
• End of chapter 4