Day 7 Worksheet.docx

Department of Computer Science and EngineeringPage 1
Exercises:SQL constraints
Question1: Considerthe following schema to
Employee (eid: integer, ename: string, age: integer, salary: real, office_location string)
A) Create a table with following constraints:
- default office_location is ‘Chandigarh’
- records of only those employees will be stored in the table who have more than 10,000
salary and minimum 25 years of age.
- Employee id and employee name should Unique values and cannot be null.
NOTE: Employee id is not a primary Key.
B) Insert values into tables confirming to constraints in question 1.
-5 records with default office location ‘Chandigarh’, 3 males and 2 females, salary between 15000 to
27000, age 25 to 35.
-5 records with office location ‘Delhi’, 2 males and 3 females, salary 45000 to 55000, age 36 to 50.
Question 2: Defining constraint names, Modifying, adding and dropping
constraints
a) Defining constraint names for constraints.
b) Change the salary check condition to 20 from 25
c) Change the salary check condition to 10000 from 20000
d) Adding unique constraint at mobile number
e) Adding default constraint on newly added department name column
f) Dropping constraints
Question 3: Create a table Employee_new with following attributes
EMPLOYEE_ID number | EMP_NAME varchar(30) | EMAIL varchar(30) | PHONE_NUMBER
number | HIRE_DATE date | GENDER char(1) | JOB_TYPE varchar(20) | SALARY number (10,2)
Create a table Employee_new with following constraints:
a) Employee_id should be an identity column generated by default with starting value of 1001 and
incremented by 1.
b) Employee name attribute should not be null.
c) Email Address attribute should be unique and not null with check constraint that email value should have
@cumail.in string at the end.
d) Phone number --not null and unique.

e) Hire date should be by default the system date.
f) Gender column should have only ‘M’ and ‘F’ values.
h) Job_type with check constraint for {‘Academics’,’Administration’, ‘Accounts’, ‘Marketing’} values.
i) Salary (with check constraint) should be between 20000 to 50000 for all employees.
Supplementary Problems:
1) For the table created in question 3 in exercises section:
a) Insert any 20 values corresponding to the attributes.
b) Assign constraint names to constraints wherever applicable.
c) Modify constraints.
d) Disable and enable different constraints.
e) Drop check constraints.
2) Create a table CLIENTS with following attributes:
Client id number, client_name varchar2(30), client_email varchar(40), order_number varchar(30), order_date
date, order_quantity number(10), shipping_date date, client_contact number(10)
With following constraints
a) Client_id should be an identity column generated by default with 100 as starting value and gets
incremented by 1.
b) Client_name should be not null
c) Client_email should be checked for @ and .characters.
d) Order number should begin with ‘O’ character followed by unique number.
e) Order_date should be system date.
f) Order_quantity should be greater than zero.
g) Shipping_date should be greater than order_date.
h) Client_contact attribute should be not null and unique.
Question 3: How to check the different constraints applied on a table using user_constraints table?
Answer 1:
CREATE TABLE Employee2 ( eid number NOTNULL,
ename VARCHAR(10) NOT NULL,
age number CHECK (age>=25),
salary number(10,2) CHECK ( salary>=10000),
OFFICE_LOCATION VARCHAR2(50) DEFAULT 'Chandigarh',

UNIQUE(eid,ename));
INSERT INTO Employee2(eid,ename,age,salary) VALUES (76,'SMIITH',24,20000);
INSERT INTO Employee2(eid,ename,age,salary) VALUES (77,'SUMIIT',25,19000);
INSERT INTO Employee2(eid,ename,age,salary) VALUES (78,'NEHAA',26,9000);
select * from Employee2;
A) Use INSERT INTO command
-
Answer 2: Defining constraint names, Modifying, adding and droping constraints
a) Defining constraint names for constraints.
CREATE TABLE Employee
( eid number CONSTRAINT cons_e NOT NULL,
ename VARCHAR(10) CONSTRAINT cons_name NOT NULL,
age number CONSTRAINT age_check CHECK (age>=25),
salary number(10,2) CONSTRAINT salary_check CHECK ( salary>=10000 ),
OFFICE_LOCATION VARCHAR2(50) DEFAULT 'Chandigarh',
CONSTRAINT uni_com UNIQUE (eid,ename))
b) Change the age check condition to 20 from 25
Drop the existing age_check constraint first using following command
ALTER TABLE employee drop constraint age_check;
Then add check constraint to age column with new condition
alter table employee modify age check(age>=20);
c) Change the salary check condition to 10000 from 20000
ALTER TABLE employee modify CONSTRAINT salary_check (salary >5000)
Gives error
Drop the existing salary_check constraint using
ALTER TABLE employee drop constraint salary_check;
Then add new constraint with constraint_name using
ALTER TABLE employee ADD CONSTRAINT salary_check CHECK (salary >20000);
d) Adding unique constraint at mobile number
Add a column contact number in the existing table using
ALTER TABLE employee add (contact number);
Then add unique constraint to this new column
ALTER TABLE employee add UNIQUE (contact);

e) Adding default constraint on newly added department name column
Add a column dept_name in the existing table using
ALTER TABLE employee add (dept_name varchar(20))
Then add default constraint to this new column
alter table employee add dept_name VARCHAR2(50) default 'CSE'
f) Dropping constraints
Drop default constraint on office_location
ALTER TABLE employee modify office_location default null;
We cannot drop the default constraint but it value can be changed to null
Dropping constraint salary check
ALTER TABLE employee_grade11 drop constraint salary_check;
Dropping unique constraint
ALTER TABLE employee_grade11 drop constraint uni_com
Question 3: Create a table Employee_new with following attributes
EMPLOYEE_ID number | EMP_NAME varchar(30) | EMAIL varchar(30) | PHONE_NUMBER number |
HIRE_DATE date | GENDER char(1) | JOB_TYPE varchar(20) | SALARY number (10,2)
Constraints:
a) Employee_id should be an identity column generated by default with starting value of 1001 and
incremented by 1.
b) Employee name attribute should not be null.
c) Email Address attribute should be unique and not null with check constraint that email value should have
@cumail.in string at the end.
d) Phone number --not null and unique.
e) Hire date should be by default the system date.
f) Gender column should have only ‘M’ and ‘F’ values.
h) Job_type with check constraint for {‘Academics’,’Administration’, ‘Accounts’, ‘Marketing’} values.
i) Salary (with check constraint) should be between 20000 to 50000 for all employees.
Answer 3:
Create table employee_new (employee_ID number GENERATED BY DEFAULT AS IDENTITY start with
100 increment by 1,
Emp_name varchar(30) NOTNULL,
Email varchar(40) NOT NULL UNIQUE check( email LIKE '%@cumail.in'), Phone_number number(10)
NOT NULL UNIQUE,

HIRE_date date default sysdate,
gender char(1) check( gender IN (‘M’,’F’),
JOB_TYPE varchar(30) check( job_type in(‘Academics’, ‘Accounts’, ‘Administration’, ‘Marketing’)),
salary number(10,2) check (salary between 20000.00 and 45000.00));
Supplementary Problems:
3) For the table created in question 3 in exercises section:
f) Insert any 20 values corresponding to the attributes.
g) Assign constraint names to constraints wherever applicable.
h) Modify constraints.
i) Disable and enable different constraints.
j) Drop check constraints.
4) Create a table CLIENTS with following attributes:
Client id number, client_name varchar2(30), client_email varchar(40), order_number varchar(30), order_date
date, order_quantity number(10), shipping_date date, client_contact number(10)
With following constraints
i) Client_id should be an identity column generated by default with 100 as starting value and gets
incremented by 1.
j) Client_name should be not null
k) Client_email should be checked for @ and .characters.
l) Order number should begin with ‘O’ character followed by unique number.
m) Order_date should be system date.
n) Order_quantity should be greater than zero.
o) Shipping_date should be greater than order_date.
p) Client_contact attribute should be not null and unique.
a) Create the table with given constraints.
CREATE TABLE client12 (
client_id NUMBER GENERATED BY DEFAULT AS IDENTITY start with 100 increment by 1,
client_name VARCHAR2(50) constraint con_name NOT NULL,
client_email varchar(40) constraint con_email NOT NULL UNIQUE check( client_email LIKE '%@%.%'),
order_number varchar(30) constraint con_order NOT NULL UNIQUE check (order_number LIKE 'O%'),
order_date date default sysdate,
order_quantity number(10) constraint con_quan check (order_quantity>0),

shipping_date date,
constraint con_shipping check(shipping_date > order_date),
client_contact number(10) constraint con_contact not null unique );
b) Insert records corresponding to attribute values.
c) Add a composite unique key constraint on email and phone number.
Alter table client12 add constraint com_uni UNIQUE( client_email, client_contact);
Question 3: How to check the different constraints applied on a table.
To check all constraints applied on a particular table then you need to query the USER_CONSTRAINTS view like
this:
SELECT * FROM user_constraints WHERE table_name = '<your table name>' AND constraint_name = '<your
constraint name>';
Select * from user_constraints where table_ name = ‘EMPLOYEE_GRADE1’
NOTE: The table name has to be specified in CAPITALS.
NOTE: In Table 1 given below only few of the columns have been shown. The next Table 2 gives the
description of all columns in the user_constraints table.
Table 1:

OWNER CONSTRAIN
T_NAME
CONSTR
AINT
_TYPE
TABLE_
NAME
SEAR
CH_C
ONDI
TION
SEARCH_CO
NDITION_VC
R_OWNER
SQL_VNWXN
PUTDLEHTCC
XPROWHQXT
T
SYS_C00285
04536
C EMPLOY
EE_GRA
DE1
EID""
IS
NOT
NULL
EID"" IS NOT
NULL
-
SQL_VNWXN
PUTDLEHTCC
XPROWHQXT
T
SYS_C00285
04537
C EMPLOY
EE_GRA
DE1
ENAM
E"" IS
NOT
NULL
ENAME"" IS
NOT NULL
-
SQL_VNWXN
PUTDLEHTCC
XPROWHQXT
T
SYS_C00285
04538
C EMPLOY
EE_GRA
DE1
age&g
t;=25
age>=25 -
SQL_VNWXN
PUTDLEHTCC
XPROWHQXT
T
SYS_C00285
04539
C EMPLOY
EE_GRA
DE1
salary
>=
10000
salary>=1
0000
-
SQL_VNWXN
PUTDLEHTCC
XPROWHQXT
T
SYS_C00285
04540
U EMPLOY
EE_GRA
DE1
- - -
Table 2:

Column Datatype NULL Description
OWNER VARCHAR2(128)
Owner of the constraint
definition
CONSTRAINT_NAME VARCHAR2(128)
Name of the constraint
definition
CONSTRAINT_TYPE VARCHAR2(1)
Type of the constraint
definition:
C - Check constraint on
a table
P - Primary key
U - Unique key
R - Referential integrity
V - With check option,
on a view
O - With read only, on a
view
H - Hash expression
F - Constraint that
involves a REF column

S - Supplemental
logging
TABLE_NAME VARCHAR2(128)
Name associated with
the table (or view) with
the constraintdefinition
SEARCH_CONDITION LONG
Text of search condition
for a check constraint.
This column returns the
correct value only when
the row originates from
the current container.
SEARCH_CONDITION_V
C
VARCHAR2(4000
)
Text of search condition
for a check constraint.
This column may
truncate the search
condition.
R_OWNER VARCHAR2(128)
Owner of the table
referred to in a
referential constraint
R_CONSTRAINT_NAME VARCHAR2(128)
Name of the unique
constraint definition for
the referenced table
DELETE_RULE VARCHAR2(9)
Delete rule for a
referential constraint:
CASCADE
SET NULL
NO ACTION

STATUS VARCHAR2(8)
Enforcement status of
the constraint:
ENABLED
DISABLED
DEFERRABLE VARCHAR2(14)
Indicates whether the
constraint is deferrable
(DEFERRABLE) or not
(NOT DEFERRABLE)
DEFERRED VARCHAR2(9)
constraint was initially
deferred (DEFERRED)
or not (IMMEDIATE)
VALIDATED VARCHAR2(13)
When STATUS =
ENABLED, possible
values are:
VALIDATED - All data
obeys the constraint
(thatis,the existingdata
in the table was
validated when the
constraint was enabled,
as well as any
subsequent data
entered into the table)

NOT VALIDATED - All
data may not obey the
constraint (that is, the
existingdatainthe table
was not validated when
the constraint was
enabled,butsubsequent
data entered into the
table was validated)
When STATUS =
DISABLED, possible
values are:
VALIDATED - All data
obeys the constraint,
but the unique index on
the constraint has been
dropped. This setting is
useful in data
warehousing
environments, but has
some restrictions. Refer
to Oracle Database Data
Warehousing Guide for
more information on
this setting.
NOT VALIDATED - All
data may not obey the
constraint

GENERATED VARCHAR2(14)
name of the constraint
is user-generated
(USER NAME) or
system-generated
(GENERATED NAME)
BAD VARCHAR2(3)
Indicates whether this
constraint specifies a
centuryinan ambiguous
manner (BAD) or not
(NULL). To avoid errors
resulting from this
ambiguity, rewrite the
constraint using the
TO_DATE function with
a four-digit year.
See Also: the TO_DATE
function in Oracle
DatabaseSQL Language
Reference and Oracle
Database Development
Guide
RELY VARCHAR2(4)
When VALIDATED =
NOT VALIDATED, this
column indicates
whether the constraint
is to be taken into
account for query
rewrite (RELY) or not
(NULL).
When VALIDATED =
VALIDATED, this
column is not
meaningful.

LAST_CHANGE DATE
When the constraint
was last enabled or
disabled
INDEX_OWNER VARCHAR2(128)
Name of the user
owning the index
INDEX_NAME VARCHAR2(128)
Name of the index (only
shown for unique and
primary-keyconstraints)
INVALID VARCHAR2(7)
constraint is invalid
(INVALID) or not (NULL)
VIEW_RELATED VARCHAR2(14)
constraint depends on a
view (DEPEND ON
VIEW) or not (NULL)
ORIGIN_CON_ID VARCHAR2(256)
The ID of the container
where the data
originates. Possible
values include:
0: This value is used for
rows in non-CDBs. This
value is not used for
CDBs.
n: This value is used for
rows containing data
that originate in the
containerwithcontainer
ID n (n = 1 if the row
originates in root)

EXERCISE QUESTIONS:Primary Key and ForeignKey
The following relational schema is given:
Emp(eid: integer, ename:string, age:integer, salary:real)
Works(eid:integer, did: integer, project_name:string)
Dept(did: integer, dname: string, budget: real, managerid: integer)
Ques 1: For basedon the given relationalschema.
a) Create emp, dept and works table with the possible primary and foreign key constraints.
CREATE TABLE Emp( eid INTEGER,
ename varCHAR(30),
age INTEGER ,
salary REAL,
constraint cons_emp_pk
PRIMARY KEY (eid))
CREATE TABLE Dept(
did INTEGER,
dname varchar2(30),
buget REAL,
managerid INTEGER ,
constraint cons_dept_pk
PRIMARY KEY (did),
constraint cons_dept_fk

FOREIGN KEY (managerid) REFERENCES Emp (eid) on delete cascade)
CREATE TABLE Works (
eid INTEGER,
did INTEGER,
project_name varchar(20),
constraint cons_works_fk
FOREIGN KEY (did) REFERENCES Dept (did) on delete set null
constraint cons_works_fk1
FOREIGN KEY(eid) REFERENCES emp(eid) on delete cascade
insert into emp values(101,'John Doe',32,15000);
insert into emp values(102,'Toy',32,15000);
insert into emp values(103,'Ralph',35,35000);
insert into emp values(104,'Joe',37,50000);
insert into emp values(105,'William',40,75000);
select * from emp;
insert into Dept values(11,'CSE',50000,101);
insert into dept values (12,'ME',100000,102);
insert into dept values (13,'ECE',75000,103);
select * from dept
insert into works values(101,11,'Research');
insert into works values (102,11,'Academics');
insert into works values (103,12,'Administration');

insert into works values (104,13,'Administration');
insert into works values (105,12,'Research');
select * from works order by eid
b) Delete a record from works table with eid = 105 and check whether referential integrity constraint
violates or not.
delete from works where eid = 105 -- eid entry corresponding to 105 gets deletedfrom works table and no
effect on parent table emp as works is the child table
c) Delete a record from emp table with eid = 104 and check whether referential integrity constraint
violates or not.
delete from emp where eid = 104 – eid entry corresponding to 104 gets deleted from both parent and child
table as a result of ‘on delete cascade’
d) Delete a record from dept table with did = 12 and check whether referential integrity constraint
violates or not.
delete from dept where did = 12 – this will remove the corresponding value from dept table and set value to ‘–‘
i.e. null for did column in works table as a result of ‘on delete set null’
EI
D
DI
D
PROJECT_NAM
E
10
3
- Administration
10
1
11 Research
10
2
11 Academics

e) Changing on delete set null to on delete NO ACTION
- NO ACTION cannot be used in the script to define a NO ACTION constraint in Oracle.
- By default if on delete clause is not mentioned in foreign key constraint then it represents no action.
Firstly drop the existing constraint using
-alter table works drop constraint cons_works_fk
Secondly, add the constraint without on delete clause which means NO ACTION Referential action
- alter table works add constraint cons_works_fk FOREIGN KEY(did) REFERENCES
dept(did)
Now, execute the following query
o delete from dept where did = 11
The following error is generated
ORA-02292: integrity constraint
(SQL_GVLIADHQLTEEFZAHBPGHKPSVD.CONS_WORKS_FK) violated - child
record found ORA-06512: at "SYS.DBMS_SQL", line 1721
NOTE: ON DELETE SET DEFAULT, ON UPDATE CASCADE, ON UPDATE SET NULL, ON UPDATE
SET DEFAULT are not supported by ORACLE
Question 2: How to define self referential integrity constraint.
Solution: Self referential integrity constraint - In this type of referential integrity, foreign key references the
primary key of same table.

For example we have employee table with attributes emp_id, emp_name, job_title and manager_id. As
manager has to be an employee with some emp_id so Manager_id is self referring the emp_id key in the same
table.
create table employee(
emp_id varchar(20) check(emp_id LIKE'E%'),
emp_name varchar(20) not null,
job_title varchar(20) not null,
manager_id varchar(20) check(manager_id LIKE 'E%'),
Constraint cons_pk PRIMARY KEY (emp_id),
Constraint cons_fk Foreign key(manager_id) references employee(emp_id))
insert into employee values('E101', 'Akash', 'Manager', 'E101')
insert into employee values('E102', 'Ridhima', 'Executive', 'E101')
insert into employee values('E103', 'Vansh', 'Manager', 'E103')
insert into employee values('E104', 'Rahul', 'Executive', 'E101')
insert into employee values('E105', 'Poorva', 'Senior_executive', 'E103')
insert into employee values('E106', 'Rohandeep', 'Senior_Executive', 'E103')
select * from employee order by emp_id
EMP_I
D
EMP_NA
ME
JOB_TITLE
MANAGER_I
D
E101 Akash Manager E101
E102 Ridhima Executive E101
E103 Vansh Manager E103
E104 Rahul Executive E101
E105 Poorva
Senior_executi
ve
E103
E106 Rohandeep
Senior_Executi
ve
E103
Now, execute following queries

1. insert into employee values('E107', 'Riva', 'Manager', 'E108') --- it gives integrity constraint violation
error as manager_id 108 is not a valid emp_id.
2. delete from employee where emp_id = 'E104' – deletes row successfully
3. delete from employee where emp_id = 'E101' --- it gives referential integrity constraint error
4. To delete an employee who is also the manager, change its manages id to null and then delete the
corresponding entry
update employee set manager_id = null where manager_id = 'E101'
select * from employee
EMP_I
D
EMP_NA
ME
JOB_TITLE
MANAGER_I
D
E101 Akash Manager -
E102 Ridhima Executive -
E104 Rahul Executive -
E105 Poorva
Senior_executi
ve
E103
E106 Rohandeep
Senior_Executi
ve
E103
Delete from employee where emp_id = ‘E101’ –deleted successfully
5. Add self referential integrity constraint with ON DELETE CASCADE
Drop table employee
Create table with on delete cascade referential action
create table employee(
emp_id varchar(20) check(emp_id LIKE'E%'),
emp_name varchar(20) not null,
job_title varchar(20) not null,

manager_id varchar(20) check(manager_id LIKE 'E%'),
Constraint cons_pk PRIMARY KEY (emp_id),
Constraint cons_fk Foreign key(manager_id) references employee(emp_id) on delete cascade)
insert into employee values('E101', 'Akash', 'Manager', 'E101')
insert into employee values('E102', 'Ridhima', 'Executive', 'E101')
insert into employee values('E103', 'Vansh', 'Manager', 'E103')
insert into employee values('E104', 'Rahul', 'Executive', 'E101')
insert into employee values('E105', 'Poorva', 'Senior_executive', 'E103')
insert into employee values('E106', 'Rohandeep', 'Senior_Executive', 'E103')
delete from employee where emp_id = 'E101' – executes successfully
EMP_I
D
EMP_NA
ME
JOB_TITLE
MANAGER_I
D
E105 Poorva
Senior_executi
ve
E103
E106 Rohandeep
Senior_Executi
ve
E103
Ques 3: The following table has two attributes A and C where A is the primary key and C is the foreign key
referencing A with on-delete cascade.
A C
-----
2 4
3 4
4 3
5 2
7 2

9 5
6 4
The set of all tuples that must be additionally deleted to preserve referential integrity when the tuple (2,4) is
deleted is:
Ans: (5,2), (7,2) and (9,5) as on deletion of tuple 2,4 leads to deletion of 5,2 and 7,2 and (5,2) deletion leads to
deletion of 9,5.
CREATE TABLE test_fk
(
A number PRIMARY KEY,
C number,
FOREIGN KEY(C) REFERENCES test_fk(A) on delete cascade)
insert into test_fk values (2,2)
update test_fk set c=4 where a = 2;
update test_fk set c=4 where a = 3;
select * from test_fk order by a
delete from test_fk where a = 2 and c = 4 – rows deleted are (2,4), (5,2),(7,2) and (9,5) will be deleted
A C
3 4
4 3
6 4

Ques 4: In table T1, P is the primary key, Q is the foreign key referencing R in table T2 with on-delete cascade
and on-update cascade. In table T2, R is the primary key and S is the foreign key referencing P in the table T1
with on-delete set NULL and on-update cascade. In order to delete record (3,8) from table T1, (8,3) T2, (5) T1,
(2) T1 numbers of additional record that need to be deleted from table T1 is:
Ans:Given,
● Q -> R(Primary Key)
● S -> P (Primary Key)
Entry to be deleted – P (3) and Q(8)
● Q can be deleted directly
● Now, S – > P but the relationship given is on delete set NULL, Therefore when we delete 3 from T1
,the entry in T2 having 3 will be NULL.
Therfore, Answer is 0 entries.
Kindly check entries corresponding to (8,3) T2, (5) T1, (2) T1

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