Rdbms Lc Labguide

RDBMS – Exercises for Hands on

ER/CORP/CRS/DB07/004 Education & Research Department© Infosys Technologies Limited 1 of 31

RDBMSExercises for Hands-on

© Infosys Technologies LtdNo. 350, Hebbal Electronics City, Hootagalli

Mysore – 571186

Author(s) Anubhav Pradhan, Umesha Murthy, Hanumesh V.J, Seema Acharya

Authorized by

Dr. M.P. Ravindra

Creation Date


BackgroundThis document contains the assignments to be completed as part of the hands on for the subject RDBMS (Course code: DB07).

Note: All assignments in this document must be completed in the sequence in this document in order to complete the course.

Day 1

Assignment 1

Draw an ER diagram to capture the requirements as stated below:A Company has several business units. Each business unit has multiple projects. Employees must be assigned to one business unit. One or more employees are assigned to a project, but an employee may be on vacation and not assigned to any projects. One of the assigned employees will be project manager for that project.

Assignment 2

Draw an ER diagram to capture the requirements as stated below:In a hospital there are different departments. Patients are treated in these departments by the doctors assigned to patients. Usually each patient is treated by a single doctor, but in rare cases they will have two or three. Healthcare assistants will also attend to patients; every department has many healthcare assistants. Each patient is required to take a variety of drugs during different parts of the day such as morning, afternoon and night.

Assignment 3

Draw an ER diagram to capture the requirements as stated below:A toy manufacturing company manufactures different types of toys. The company has several manufacturing plants. Each plant manufactures different types of toys. A customer can place the order for these toys. Each order may contain one or more toys. Each customer has multiple ship-to addresses. To promote the business, the company offers different schemes based on the order value.



Day 2

Assignment 4

Convert the ER model into a relational Schema



Assignment 5Let us find out the Primany Key and highest normal form for this relation:

R{a,b,c,d,e,} a,e c, d e, a,eb, a,ed

Solution:

Determining primary key : From above stated functional dependencies, it is evident that a,e determines c and d . d determines e and hence a,e determines all the non-key attributes (b,c,d) of the relation. Thus a,e is the primary key for relation R.

Highest normal form: All the attributes are atomic in nature therefore Relation R is in 1NF.

All non-key attributes (b,c,d) are fully functionally dependent on the primary key (a,e). Therefore the relation in is 2NF.

All the non-key attributes (b,c,d) are non-transitively depending on primary key (a,e). Therefore the relation is in 3NF.

e is a part of the primary key, therefore its dependence on d does not violate 3NF definition.

Relation R is not in BCNF because d determines e and d, although a determinant is not a candidate key.

Assignment 6

Find the primary key and the highest normal form of following relations

a. R{a,b,c} a,bc and ca

b. R{a,b,c,d} ab, ad and bc

c. R{a,b} ab and ba

d. R{a,b,c,d,e} a,bc, a,bd and de



Assignment 7Convert following table to 3NF table. Please add appropriate columns in normalized tables to make 3NF compliant.

CustomerID

Customer_Name

Customer_Address

Product_Name

Product_Brand

Product_Category

Sell Price

Date_of_Transaction

Qty

101 Lawrence21, Lion Blvd, St Peter Excel Surf

Detergent and Powder 43 11-Feb-05 2

102 Jagan29, Tiger Blvd, Burbank GlucoBis Goodday

Cookies and Confectionaries 8 22-Mar-05 12

103 Bobby213,Parliament Blvd, Petas Rotomax KingTime

Watch and clocks 120 27-Mar-05 8

104 Ramsey401,Gazipura Blvd,Gulbarga SouthernStar CoolPlus Garments 63 5-Apr-05 3

105 Chang Hu101,Marshal Blvd,Kingston Winner Fusion Stationary 3 22-Apr-05 27

Assignment 8

Draw the E-R diagram for above normalized Relational schema.



Day 3

Assignment 9Configuring ORACLE net service on your machineObjective: To learn how to create host string using net service to connect oracle client to the oracle server.Background: Although on your machine net service is pre configured but please follow these steps to get a feel of how actually this has been configured.

Step 1:

Reach OracleNet8 Assistant via :Start->All programs->oracle-orahome81->networkadministration->net8assistant



Step2:

You will get an interface like Click on service Naming and click the “+” button to the left



Step 3:

After clicking + wizard appears

Enter Net Service Name as training (If it already exists chose any other name)



Step 4:

Click the next buttonChoose TCP/IP as the protocol and click next

Step 5:

Enter Host Name given to you and don’t change port number (default is 1521)



Step 6:

Click next

Choose (oracle8i) and service name is “training”

Step 7:

Test the connection by clicking the test button



Default it will test with the userid as “scott” and password as “tiger”,but you give your UserID and password and test that also



Click on change login button and give your userid and password

For example userid is trng180 password is “Infosys”

Once again test the connection after giving username and password



Now configuration is tested

Step 8:

After that login into SQL plus by following the set of links as given below

Enter your username ex trng180 password is Infosys and host string is training



you will get SQL prompt and start working Summary of this exercise:You have just learnt

Net 8 is the service for communication between oracle client and server How to configure Net service.



Assignment 10Solving simple queries

Create two tables - EMP & DEPT

EMPColumn name

Data type Description

EMPNO Number Employee numberENAME Varchar Employee nameJOB Char DesignationMGR Number Manager’s Emp.

numberHIREDATE Date Date of joiningSAL Number Basic SalaryCOMM Number CommissionDEPTNO Number Department Number

DEPTColumn name

Data type Description

DEPTNO Number Department numberDNAME Varchar Department nameLOC Varchar Location of

department

Data for EMP7369 Smith Clerk 7902 17/12/8

0800 20

7499 Allen Salesman

7698 20/2/81 1600 300 30

7521 Ward Salesman

7698 22/2/81 1250 500 30

7566 Jones Manager 7839 2/4/81 2975 207654 Martin Salesma

n7698 28/9/81 1250 1400 30

7698 Blake Manager 7839 1/5/81 2850 307782 Clark Manager 7839 9/6/81 2450 107788 Scott Analyst 7566 9/12/82 3000 207839 King President 17/11/8

15000 10

7844 Turner Salesman

7698 8/9/81 1500 0 30

7876 Adams Clerk 7788 12/1/83 1100 207900 James Clerk 7698 3/12/81 950 307902 Ford Analyst 7566 4/12/81 3000 20



7934 Miller Clerk 7782 23/1/82 1300 10

Data for DEPT table10 Accounting New York20 Research Dallas30 Sales Chicago40 Operations Boston

Assignment 11

Perform the following queries on the EMP and DEPT table:

a) List the names of analysts and salesmen.

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

c) List names of employees who are not managers.

d) List the names of employees whose employee numbers are 7369, 7521, 7839, 7934, 7788.

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

f) List employee names for those who have joined between 30 June and 31 Dec. ‘81.

g) List the different designations in the company.

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

i) List the name and designation of the employee who does not report to anybody.

j) List the employees not assigned to any department.

k) List the employees who are eligible for commission.

l) List employees whose names either start or end with “S”.

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

n) List the number of employees working with the company.

o) List the number of designations available in the EMP table.



p) List the total salaries paid to the employees.

q) List the maximum, minimum and average salary in the company.

r) List the maximum salary paid to a salesman.



Assignment 12

Understanding a simple query

Objective: To visualize how a simple query works.

Problem Statement: Let us try to understand how the following query works:

“List employees not belonging to department 30, 40, or 10”

The solution is:

SELECT ename FROM emp WHERE deptno NOT IN (30, 40, 10);

Let us explore how this query is working.

Step 1: The query filters those tuples which do not belong to deptno 30, 40 or 50, because of the operator NOT IN, from the emp table.

Step 2: Finally it displays only the employee names of the filtered tuples because we specify only ename in the SELECT clause.

Summary of this exercise:You have just learnt

How a simple query works.



Day 4

Assignment 13Objective: To visualize how group by and having works.Problem Statement: Let us try to understand how the following query works:

“List the total salary, maximum and minimum salary and average salary of the employees jobwise, for department 20 and display only those rows having an average salary > 1000”

The solution is:

SELECT job, avg(sal) from emp where deptno=20 group by job having avg(sal) > 1000 order by job;Let us explore how this query is working.

Note: All columns in the column clause of your select statement that are not in a group function must be listed in the group by clause. However a column listed in the group by clause needn’t appear in the column clause.

Step 1: First thing first we have to select (filter) data only for dept 20. For this we have to put

WHERE deptno=20Step 2: Now, we have to find average salary of the employees job wise. Therefore we have to group our selected (filtered) data using group by clause.

GROUP BY jobStep 3: Now, we have to display only those rows having an average salary > 1000. Therefore we have to put

HAVING avg(sal) > 1000Note: When you are using where, group by and having together than the order should be as follows: where, group by and having.

Step 4: For getting the final output in ascending order of job we have to put order by clause.

ORDER BY jobNote: Order by clause should be the last clause in the query otherwise you will get the error as follows (In oracle): ORA-00933: SQL command not properly ended.

Summary of this exercise:You have learnt

Where to use where and having and there differences.



The order of group by, having and where. The placement of order by.

Assignment 14

14. Perform the following queries against EMP and DEPT tables:

Note: Use sysdate to get the system’s date. For example : select sysdate from emp;

a) List the number of employees and average salary for employees in department 20.

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

c) List names of employees who are more than 2 years old in the company. d) List the employee details in the ascending order of their basic salary.e) List the employee name and hire date in the descending order of the hire

date.f) 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.

g) List the department numbers and number of employees in each department.h) List the department number and total salary payable in each department.i) List the jobs and number of employees in each job. The result should be in

the descending order of the number of employees.j) List the total salary, maximum and minimum salary and average salary of the

employees jobwise.k) List the total salary, maximum and minimum salary and average salary of the

employees, for department 20.l) List the average salary of the employees job wise, for department 20 and

display only those rows having an average salary > 1000



Assignment 15

The following questions pertain to a database with the following tables.Suppliers

S# SName

Status City

S1 Smith 20 LondonS2 Jones 10 ParisS3 Blake 30 ParisS4 Clark 20 LondonS5 Adams 30 Athens

PartsP# PNam

eColor Weight City

P1 Nut Red 12.0 LondonP2 Bolt Green 17.0 ParisP3 Screw Blue 17.0 RomeP4 Screw Red 14.0 LondonP5 Cam Blue 12.0 ParisP6 Cog Red 19.0 London

ProjectsJ# Jname CityJ1 Montago LondonJ2 Cat ParisJ3 Box LondonJ4 Montago RomeJ5 Eagles Athens

ShipmentS# P# J# QtyS1 P1 J1 350S1 P3 J3 120S2 P1 J1 620S3 P2 J3 700S2 P2 J4 250S4 P3 J2 125S5 P4 J2 325



15. 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).a) Get full details of all projects in London.

b) Get S# for suppliers who supply project J1.

c) Get all part-color/part-city combinations.

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

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

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

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

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

i) Get all pairs of part numbers such that some supplier supplies both the indicated parts.

j) Get the total quantity of part P1 supplied by S1.

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

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

m)Get project names for projects supplied by supplier S1.

n) Get colors of parts supplied by S1.

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

p) Get supplier numbers for suppliers supplying at least one part supplied by at least one supplier who supplies at least one red part.

q) Get supplier numbers for suppliers with a status lower than that of supplier S1.

r) Get project numbers for projects not supplied with any red part by any London supplier.



Assignment 16

Objective: To visualize how an inner query works.

Problem Statement: Let us consider the following query:

“Get supplier numbers for suppliers with a status lower than that of supplier S1”

The solution is:

SELECT S# FROM Suppliers WHERE Status < (SELECT Status FROM Suppliers WHERE S#=’S1’);


Step 1: The inner query is executed first which selects the status of supplier s1 from suppliers table.

Step 2: Now the query looks like this:

SELECT S# FROM Suppliers WHERE Status < 5

(Considering the status of S1 is 5).

Step 3: Above query is again executed as a simple query i.e. tuples are selected for those suppliers whose status is less than 5 and finally there s# is being displayed.


Inner query executes first and once only.



Day 5

Assignment 17Objective: To visualize how correlated query works. Background: Let us consider the following query:

“List drivers who have delivered shipments to every city”

The solution is:

SELECT a.driver_name FROM truck a WHERE NOT EXISTS (SELECT * FROM city b WHERE NOT EXISTS(

SELECT * FROM shipment c where c.destination=b.city_name and a.truck# = c.truck#));


Step 1: The outermost query is executed first and for the first driver_name in truck relation NOT EXISTS condition is being verified. Step 2: Against the first driver_name all the cities are selected and again one by one for each city shipment destination is being checked with the city_name in the innermost query.

NOTE: This is a correlated query because the innermost query is referring to a column in the next outer query.

Step 3: If the condition c.destination=b.city_name become false for any of the tuple than NOT EXISTS returns false and hence that city and henceforth that driver is not being considered in the final output.

Step 4: Step 1 to step3 will be repeated for all the turck drivers in the truck relation.


Working of NOT EXISTS. A sample correlated query.



Assignment 1818. Now solve the following set of queries according to given schema:

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 10,000ANNUAL_REVENUE defaults to $20,000CUST_TYPE must be manufacturer, wholesaler, or retailer

Data for CUSTOMER table

CUST_ID

CUST_NAME

ANNUAL_REVENUE CUST_TYPE

100 Revathi 1000000 manufacturer101 Richa 1800000 wholesaler102 Rishi 1000000 retailer103 Rijesh 4000000 wholesaler104 Kalyan 4800000 Wholesaler311 Karthik 5500000 retailer

SHIPMENT (SHIPMENT_#, CUST_ID, WEIGHT, TRUCK_#, DESTINATION, SHIP_DATE)Foreign Key: CUST_ID REFERENCES CUSTOMER, on deletion cascadeForeign Key: TRUCK_# REFERENCES TRUCK, on deletion set to nullForeign Key: DESTINATION REFERENCES CITY, on deletion set to nullWEIGHT must be under 1000 and defaults to 10

Data for SHIPMENT table

SHIPMENT_#

CUST_ID

WEIGHT

TRUCK_#

DESTINATION

SHIP_DATE

100 100 500 100 London Null101 101 100 102 Paris Null102 101 300 103 London Null103 101 10 102 Panama City 12-Dec-03104 101 20 101 Los Angeles Null105 102 200 102 Rome Null106 100 50 101 Sioux City 18-Sep-03107 104 500 100 Manhattan Null108 103 50 103 San

FranciscoNull

109 104 25 101 San Francisco

Null



110 102 200 103 London 11-Oct-98111 103 100 101 London 09-Sep-99112 104 500 100 London 18-Jun-88113 104 200 100 London 11-Oct-98114 104 50 103 Manhattan 29-May-03115 100 75 103 Los Angeles 17-Sep-02116 101 55 102 Baltimore 01ul-02117 103 45 101 Paris Null118 103 45 100 Rome Null119 103 45 102 Los Angeles Null120 104 45 102 London Null121 100 150 102 Sioux City Null122 101 500 102 Manhattan Null123 102 250 102 San

Francisco31-Jul-02

124 311 0.5 102 Denver Null125 311 100 102 St. Louis Null

TRUCK (TRUCK_#, DRIVER_NAME)

Data for TRUCK table

TRUCK_#

DRIVER_NAME

100 Jensen101 Sasi102 Hrithik103 Jake Stinson

CITY (CITY_NAME, POPULATION)

Data for CITY table

CITY_NAME POPULATION

London 100000000Paris 120000000Rome 200000000Panama City 1230000000San Francisco

20000000

Sioux City 5000000000Manhattan 10000000Los Angeles 7000



Baltimore 2000Denver 1000St. Louis 5000

Perform the following queries:

a) What are the names of customers who have sent packages (shipments) to Sioux City?

b) To what destinations have companies with revenue less than $1 million sent packages?

c) What are the names and populations of cities that have received shipments weighing over 100 pounds?

d) Who are the customers having over $5 million in annual revenue who have sent shipments weighing less than 1 pound?

e) 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?

f) Who are the drivers who have delivered shipments for customers with annual revenue over $20 million to cities with populations over 1 million?

g) List the cities that have received shipments from customers having over $15 million in annual revenue.

h) List the names of drivers who have delivered shipments weighing over 100 pounds.

i) List the name and annual revenue of customers who have sent shipments weighing over 100 pounds.

j) List the name and annual revenue of customers whose shipments have been delivered by truck driver Jensen.

k) List customers who had shipments delivered by every truck. ( use NOT EXISTS)

l) List cities that have received shipments from every customer. ( use NOT EXISTS)

m) List drivers who have delivered shipments to every city. (use NOT EXISTS)n) Customers who are manufacturers or have sent a package to St. Louis.o) Cities of population over 1 million which have received a 100-pound package

From customer 311.p) Trucks driven by Jake Stinson which have never delivered a shipment to

Denver.q) Customers with annual revenue over $10 million which have sent packages

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

r) 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.



s) 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?



Day 6

Assignment 19Objective: To learn about the exclusive lock. Background: The different locking mechanisms have been explained to you in the class. For practice, we will use EMP table (which you have used in Assignments for Day3).

Note: In Oracle the default locking is row exclusive lock for Update/Delete/Insert

Step 1: Open two instances of SQLPLUS on your machine and login with your Oracle ID in both the instances.

Step 2: Go to first instance.Write the following query:

update emp set sal=9000 where empno=7369;

Step 3: Go to the second instance.Write the following query:


Note: The second instance is hanged. The reason is the first transaction has issued the update statement for the empno 7369 and it acquires the exclusive(X) lock. The second transaction is also trying to update the same but because of the X lock acquired by the first transaction it has to wait.

Step 4: Go to the first instance and issue COMMIT/ROLLBACK.

Step 5: Go to the second instance. You get the SQL prompt.Note: As soon as you Commit/Rollback the transaction in second instance the X lock is released.

Step 6: Issue COMMIT/ROLLBACK in the second instance as well.


Update/Delete/Insert DML statement acquire a row exclusive lock



The X lock is released only at the end of transaction which is marked by Commit/Rollback.

Assignment 20

Objective: To visualize how deadlock occurs.

Step 1: Open two instances of SQLPLUS (If you have closed the previous ones) on your machine and login with your Oracle ID in both the instances.







Note: The first instance is hanged. What is the reason? Do you remember the last hands on?



Note: This situation is deadlock. Oracle is smart enough and detect the same and it rollbacks the statement of Step 4. Check the Oracle message: ORA-00060: deadlock detected while waiting for resource

Step 6: Issue COMMIT/ROLLBACK in both the instances.


How to visualize the deadlock.



Oracle intelligently determines the deadlock situation


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Rdbms Lc Labguide