Assignment Solutions for RDBMS Punjabi University … · Assignment Solutions for RDBMS Punjabi...

Assignment Solutions for RDBMS Punjabi University Patiala

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Assignments Day 3

1. Create a database called COMPANY consisting of two tables - EMP &

DEPT


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2. Perform the following queries on the tables just created:

1. List the names of analysts and salesmen.

SQL> select ename from emp where job=‟analyst‟ and job=‟salesman‟;

2. List details of employees who have joined before 30 Sep 81.

SQL> select * from emp where hiredate < ‟30-sep-81‟;

3. List names of employees who are not managers.

SQL> select ename from emp where job is not manager;

4. List the names of employees whose employee numbers are 7369, 7521, 7839,

7934, 7788.

SQL> select ename from from emp where empno in

(7369,7521,7839,7934,7788);

5. List employees not belonging to department 30, 40, or 10.

SQL> select ename from emp where deptno not in (30,40,10);

6. List employee names for those who have joined between 30 June and 31

Dec. 81.

SQL> select ename from emp where hiredate between ‟30-jun-81‟ and ‟31-

dec-81‟;

7. List the different designations in the company.

SQL> select distinct job from emp;

8. List the names of employees who are not eligible for commission.

SQL> select ename from emp where comm = NULL;

9. List the name and designation of the employee who does not report to

anybody.

SQL> select ename,job from emp where job = „president‟;


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10. List the employees not assigned to any department.

SQL> select ename from emp where job = NULL;

11. List the employees who are eligible for commission.

SQL> select ename from emp where comm. is not NULL;

12. List employees hose names either start or end with “S”.

SQL> select ename from emp where ename like „S%‟ and like „%S‟;

13. List names of employees whose names have “i” as the second character.

SQL> select ename from emp where ename like „_i%‟;

14. List the number of employees working with the company.

SQL> select ename from emp;

15. List the number of designations available in the EMP table.

SQL> select distinct job from emp;

16. List the total salaries paid to the employees.

SQL> select sum(sal) from emp;

17. List the maximum, minimum and average salary in the company.

SQL> select max(sal),min(sal),avg(sal) from emp;

18. List the maximum salary paid to a salesman.

SQL> select max(sal) from emp where job = „salesman‟;


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Assignment Day 4

1) Please refer to the tables created as a part of Assignment 3. Perform the

following queries against those tables:

1. List the number of employees and average salary for employees in

department 20.

SQL> select count(ename),avg(sal) from emp where deptno = 20;

2. List name, salary and PF amount of all employees. (PF is calculated as 10%

of basic salary)

SQL> select ename,sal,sal((sal/100)*10) from emp;

3. List names of employees who are more than 2 years old in the company.

SQL>

4. List the employee details in the ascending order of their basic salary.

SQL> select * from emp order by sal;

5. List the employee name and hire date in the descending order of the hire

date.

SQL> select ename,hiredate from emp order by hiredate desc;

6. List employee name, salary, PF, HRA, DA and gross; order the results in the

ascending order of gross. HRA is 50% of the salary and DA is 30% of the

salary.

SQL> select ename,sal, ((sal/100)*10)”PF”, ((sal/100)*50)”HRA”,

((sal/100)*30)”DA”,

(sal+((sal/100)*10)+((sal/100)*50)+((sal/100)*30))”Gross” from emp;

7. List the department numbers and number of employees in each department.

SQL> select deptno,count(ename) from emp group by deptno;


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8. List the department number and total salary payable in each department.

SQL> select deptno,sum(sal) from emp group by deptno;

9. List the jobs and number of employees in each job. The result should be in

the descending order of the number of employees.

SQL> select job,count(empno) from emp group by job;

10. List the total salary, maximum and minimum salary and average salary of

the employees jobwise.

SQL> select sum(sal),max(sal),min(sal),avg(sal) from emp group by job;


the employees, for department 20.

SQL> select sum(sal),max(sal),min(sal),avg(sal) from emp where deptno = 20

group by job;


the employees jobwise, for department 20 and display only those rows

having an average salary > 1000

SQL> select sum(sal),max(sal),min(sal),avg(sal) from emp where deptno = 20

having avg(sal) > 1000 group by job;


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2) The following questions pertain to a database with the following tables.

Suppliers - S (S#, Name, Status, City)

Parts - P (P#, Pname, Colour, Weight, City)

Projects - J (J#, Jname, City)

Shipment - SPJ (S#, P#, J#, Qty)

The significance of an SPJ record is that the specified supplier supplies the

specified part to the specified project in the specified quantity (and the

combination S#-P#-J# uniquely identifies such a record).

1. Get full details of all projects in London.

SQL> select * from J where city = „London‟;

2. Get S# for suppliers who supply project J1.

SQL> select S# from S where S# in (select S# from SPJ where J# = J1);

3. Get all part-color/part-city combinations.

SQL> select city,colour from P group by (city,colour);

4. Get all S#/P#/J# triples such that all are co-located.

SQL> select S#,P#,J# from S,P,J where S.city = P.city and P.city = J.city and

J.city = S.city and (S#,P#,J#) in (select S#,P#,J# from SPJ);

5. Get al S#, P#, J# triples such that they are not all co-located.

SQL> select S#,P#,J# from S,P,J where (S.city <> J.city or J.city <> P.city or

S.city <> P.city) and (S#,P#,J#) in (select S#,P#,J# from SPJ);

6. Get P# for parts supplied by a supplier in London.

SQL> select distinct (P#) from SPJ where S# in (select S# from A where city =

„London‟);

7. Get all pairs of cities such that a supplier in the first city supplies to a Project

in the second city.


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SQL> select S.city,J.city from S,J where S.city <> J.city and (S#,J#) in (select

S#,J# from SPJ);

8. Get J# for projects supplied by at least one supplier not in the same city.

SQL> select distinct (J#) from J where exist (select S# from S where S.city <>

J.city and (J#,S#) in (select J#,S# from SPJ));

9. Get all pairs of part numbers such that some supplier supplies both the

indicated parts.

SQL> select SPJ.P#,P.P# from SPJ, P where P.P# <> SPJ.P# and (S#,P.P#) in

(select S#,P# from SPJ) group by (SPJ.P#,P.P#);

10. Get the total quantity of part P1 supplied by S1.

SQL> select sum(qty) from SPJ where P# = „P1‟ and S# = „S1‟;

11. For each part supplied to a project, get the P#, J# and corresponding total

quantity.

SQL> select P#,J#,sum(qty) from SPJ group by (P#,J#);

12. Get P# of parts supplied to some project in an average quantity > 320.

SQL> select P#,avg(qty) from SPJ goup by P# having avg(qty)>320;

13. Get project names for projects supplied by supplier S1.

SQL> select Pname from P where P# in (select P# from SPJ where S# = „S1‟);

14. Get colors of parts supplied by S1.

SQL> select colour from P where P# in (select P# from SPJ where S# = „S1‟);

15. Get J# for projects using at least one part available from supplier S1.

SQL> select J# from J where J# in (select J# from SPJ where S# = „S1‟);


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16. Get supplier numbers for suppliers supplying at least one part supplied by at

least one supplier who supplies at least one red part.

SQL> select S# from S where S# in (select S# from SPJ where P# in (select P#

from SPJ where S# in (select S# from SPJ where P# in (select P# from SPJ

where P# in (select P# from P where colour =‟red‟)))));

17. Get supplier numbers for suppliers with a status lower than that of supplier

S1.

SQL> select S# from S where status < (select status from S where S# = „S!);

18. Get project numbers for projects not supplied with any red part by any

London supplier.

SQL> select J# from J where J# in (select J# from SPJ where P# not in (select

P# from P where colour = „red‟)) and S# in ( select S# from S where city

=‟London‟);


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Assignment Day 5

1) Write the SQL commands to create a database schema for the following

relational schema:

CUSTOMER (CUST_ID, CUST_NAME, ANNUAL_REVENUE,

CUST_TYPE)

CUST_ID must be between 100 and 10,000

ANNUAL_REVENUE defaults to $20,000

CUST_TYPE must be manufacturer, wholesaler, or retailer

SHIPMENT (SHIPMENT_#, CUST_ID, WEIGHT, TRUCK_#,

DESTINATION, SHIP_DATE)

Foreign Key: CUST_ID REFERENCES CUSTOMER, on deletion cascade

Foreign Key: TRUCK_# REFERENCES TRUCK, on deletion set to null

Foreign Key: DESTINATION REFERENCES CITY, on deletion set to null

WEIGHT must be under 1000 and defaults to 10

TRUCK (TRUCK_#, DRIVER_NAME)

CITY (CITY_NAME, POPULATION)

Perform the following queries:

1. What are the names of customers who have sent packages (shipments) to

Sioux City?

SQL> select cust_id,cust_name from customer where cust_id in (select cust_id

from shipment where designation = „Sioux‟);

2. To what destinations have companies with revenue less than $1 million sent

packages?

SQL> select distinct(designation) from shipment where cust_id in (select

cust_id from customer where annual_revenue < $1);

3. What are the names and populations of cities that have received shipments

weighing over 100 pounds?


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SQL> select city_name,population from city where city_name in (select

designation from shipment where weight > 100);

4. Who are the customers having over $5 million in annual revenue who have

sent shipments weighing less than 1 pound?

SQL> select cust_id,cust_name from customers where annual_revenue > $5

and cust_id in ( select cust_id from shipment where weight <1);

5. Who are the customers having over $5 million in annual revenue who have

sent shipments weighing less than 1 pound or have sent a shipment to San

Francisco?

SQL> select cust_id,cust_name from customers where annual_revenue > $5

and cust_id in (select cust_id from shipment where weight < 1 or designation =

‟San Francisco‟);

6. Who are the drivers who have delivered shipments for customers with

annual revenue over $20 million to cities with populations over 1 million?

SQL> select truck_#,driver_name from truck where truck_# in (select truck_#

from shipment where cust_id in (select cust_id from customer where

annual_revenue > $20) and destination in (select city_name from city where

position > 1000000));

7. List the cities that have received shipments from customers having over $15

million in annual revenue.

SQL> select distinct(designation) from shipment where cust_id in ( select

cust_id from customer where annual_revenue > $15);

8. List the names of drivers who have delivered shipments weighing over 100

pounds.

SQL> select truck_#,driver_name from truck where truck_# in (select truck_#

from shipment where weight > 100);

9. List the name and annual revenue of customers who have sent shipments

weighing over 100 pounds.


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SQL> select cust_name,annual_revenue from customer where cust_id in

(select cust_id from shipment where weight > 100);

10. List the name and annual revenue of customers whose shipments have been

delivered by truck driver Jensen.

SQL> select cust_name,annual_revenue from customer where cust_id in

(select cust_id from shipment where truck_# in (select truck_# from truck

where driver_name = „Jensen‟));

11. List customers who had shipments delivered by every truck.

SQL> select cust_id,cust_name from customer where ( select

count(distinct(truck_#)) from shipment where shipment.cust_id =

customer.cust_id) >= (select count(*) from truck);

12. List cities that have received shipments from every customer.

SQL> select city_name from city where (select count (distinct ( cust_id)) from

shipment where designation = city.city_name) >= (select count (*) from

customer);

13. List drivers who have delivered shipments to every city.

SQL> select driver_name from truck where truck_# in (select truck_# from

truck where (select count(distinct(designation)) from shipment where

shipment.truck_# = truck.truck#) >= (select count (*) from city));

14. Customers who are manufacturers or have sent a package to St. Louis.

SQL> select cust_id,cust_name from customer where cust_type =

„manufacturer‟ or cust_id in (select cust_id from shipment where designation =

„St. Louis‟);

15. Cities of population over 1 million which have received a 100-pound

package From customer 311.

SQL> select city_name from city where population > 1000000 and city_name

in (select designation from shipment where weight = 100 and cust_id = 311);


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16. Trucks driven by Jake Stinson which have never delivered a shipment to

Denver.

SQL> select truck_# from truck where driver_name = „Jake Stinson‟ and

truck_# in (select truck_# from shipment where designation <> „Denver‟);

17. Customers with annual revenue over $10 million which have sent packages

under 1 pound to cities with population less than 10,000.

SQL> select cust_id,cust_name from customer where annual_revenue >

10000000 and cust_id in (select cust_id from shipment where designation in

(select city_name from city where population < 10000) and weight < 1);

18. Create views for each of the following:

a. Customers with annual revenue under $1 million.

b. Customers with annual revenue between $1 million and $5 million.

c. Customers with annual revenue over $5 million.

a. SQL> create view revenue_1 as select cust_id,cust_name from customer

where annual_revenue < 1000000;

b. SQL> create view revenue_2 as select cust_id,cust_name from customer

where annual_revenue > 1000000 and annual_revenue < 5000000;

c. SQL> create view revenue_3 as select cust_id, cust_name from customer

where annual_revenue > 5000000;

19. Use these views to answer the following queries:

a. Which drivers have taken shipments to Los Angeles for customers with

revenue over $5 million?

b. What are the populations of cities which have received shipments from

customers with revenue between $1 million and $5 million?

c. Which drivers have taken shipments to cities for customers with revenue

under $1 million, and what are the populations of those cities?


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a. SQL> select * from revenue_3 where cust_id in (select cust_id from

shipment where designation = „Los Angeles‟);

b. SQL> select city_name,population from city where city_name in (select

designation from shipment where cust_id in (select cust_id from

revenue_2));

c. SQL> select driver_name,population from truck,city where (

truck_#,city_name) in (select truck_#, designation from shipment where

cust_id in (select cust_id from revenue_1));


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PL/SQL

PL/SQL Blocks

1. Construct a PL/SQL block to find greater of two numbers.

Declare

A number:=&enter_A;

B number:=&enter_B;

Begin

If A>B then

Dbms_output.put_line(„A is greater‟);

Else

Dbms_output.put_line(„B is greater‟);

End if;

End;


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2. Construct a PL/SQL block to find greatest of three numbers.

Declare

A number:=&enter_A;

B number:=&enter_B;

C number:=&enter_C;

Begin

If A>B then

If A>C then

Dbms_output.put_line(„A is greatest‟);

Else

Dbms_output.put_line(„C is greatest‟);

End if;

Else

If B>C then

Dbms_output.put_line(„B is greater‟);

Else

Dbms_output.put_line(„C is greater‟);

End if;

End if;

End;


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3. A PL/SQL block to calculate addition, subtraction, multiplication and

division of two numbers according to users choice.

Declare

A number:=&A;

B number:=&B;

C number;

X number;

Begin

X:=&enter_choice;

If X=1 then

C:=A+B;

Else if X=2 then

C:=A-B;

Else if X=3 then

C:=A*B;

Else if X=4 then

C:=A/B;

Else

Dbms_output.put_line(„Not a valid option‟);

End if;

Dbms_output.put_line(„Result is‟||C);

End;


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4. A PL/SQL block to print the numbers from 1 to n.

Declare

Var1 number:=1;

n number:=&enter_n;

Begin

Loop

Dbms_output.put_line(var1);

Var1:=var1+1;

Exit when var1>n;

End loop;

End;


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5. A PL/SQL block to print the desired multiplication table.

Declare

Table_of number:=&enter_tableof;

Count number:=1;

Result number;

Begin

While count<=10

Loop

Result:=table_of *count;

Dbms_output.put_line(table_of||‟*‟||count||‟=‟||result);

Count:=count+1;

End loop;

End;


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Cursor management

1. A PL/SQL code to display a message to check whether the record is deleted

or not.

Begin

Delete emp where empno=&empno;

If sql%notfound then

Dbms_output.put_line(„record not deleted‟);

Else

Dbms_output.put_line(„record deleted‟);

End if;

End;


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2. A PL/SQL code to display a message to give the number of records deleted

by the delete statement issued in a PL/SQL block.

Declare

n number;

Begin

Delete emp where deptno=&deptno;

N:=sql%rowcount;

Dbms_output.put_line(„total number of record deleted‟||n);

End;


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3. A PL/SQL code to display the empno, ename, job of employees of

department number 10.

Declare

Cursor c1 is select empno, ename, job from emp where deptno=10;

Rec c1%rowtype;

Begin

Open c1;

Loop

Fetch c1 into rec;

Exit when c1%notfound;

Dbms_output.put_line(„empno‟||rec.empno);

Dbms_output.put_line(„ename‟||rec.ename);

Dbms_output.put_line(„job‟||rec.job);

End loop;

Close c1;

End;


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4. A PL/SQL code to display the employee number and name of top 5 highest

paid employees.

Declare

Empname emp.ename%type;

Empsal emp.sal%type;

Cursor temp1 is select ename, sal from emp order by sal desc;

Begin

Open temp1;

Loop

Fetch temp1 into empname, empsal;

Exit when temp1%rowcount>5 or temp1%notfound;

Dbms_output.put_line(empname||empsal);

End loop;

Close temp1;

End;


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5. A PL/SQL code to calculate the total salary of first n records of emp table.

The value of n is passed to cursor as parameter.

Declare

e number(5);

s number(5):=0;

cursor c1(n number) is select sal from emp where rownum<=n;

Begin

Open c1(&n);

Loop

Fetch c1 into e;

Exit when c1%notfound;

S:=s+e;

End loop;

Dbms_output.put_line(s);

Close c1;

End;


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Triggers

1. To create a trigger for the emp table, which makes the entry in ename

column in uppercase.

Create or replace trigger upper_trigger

Before insert or update of ename on emp

For fetch row

Begin

:new.ename:=upper(:new.ename);

End;


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2. A trigger on emp table, which shows the old values and new values of

ename after any updation on ename of emp table.

Create or replace trigger emp_update

After update of ename on emp for each row

Begin

Dbms_output.put_line(„old name:‟||:old.ename);

Dbms_output.put_line(„new name:‟||:new.ename);

End;


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3. A trigger on the emp table, which store the empno and operation in table

auditor for each operation i.e. insert, update and delete.

Create or replace trigger emp_audit

After insert or update on emp

For each row

Begin

If inserting then

Insert into auditor values(:new.empno,‟insert‟);

Elsif updating then

Insert into auditor values(:new.empno,‟update‟);

Elsif deleting then

Insert into auditor values(:new.empno,‟delete‟);

End if;

End;


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4. To create a trigger so that no operation can be performed on emp table on

Sunday.

Create or replace trigger emp_Sunday

Before insert or update or delete on emp

Begin

If rtrim(upper(to_char(sysdate,‟day‟)))=‟sunday‟ then

Raise_application_error(-20022, „no operation can be performed o

Sunday‟);

End if;

End;


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5. To create a trigger, which verify that updated salary of employee must be

greater than his/her previous salary.

Create or replace trigger update_check

Before update on emp

For each row

Begin

If :new.sal<:old.sal then

Raise_application_error(-20000. „new salary cannot be less than old

salary‟);

End if;

End;

Assignment Solutions for RDBMS Punjabi University … · Assignment Solutions for RDBMS Punjabi...

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