BITS WILP Database Design and Applications Mid-Sem Exam 2016-H2 (Regular)



Birla Institute of Technology & Science, Pilani
Work-Integrated Learning Programmes Division
First Semester 2016-2017

Mid-Semester Test
(EC-2 Regular)

Course No.                  : SS ZG518  
Course Title                 : DATABASE DESIGN AND APPLICATIONS  
Nature of Exam           : Closed Book
Weightage                    : 35%
Duration                      : 2 Hours 
Date of Exam              : 25/09/2016    (AN)
No of pages: 2
No of questions: 4

Note:
1.       Please follow all the Instructions to Candidates given on the cover page of the answer book.
2.       All parts of a question should be answered consecutively. Each answer should start from a fresh page. 
3.       Assumptions made if any, should be stated clearly at the beginning of your answer.

Q.1 (a)          Mention the 3 basic components of  E-R model                                                       [1.5]

E-R Model components
1)      Entity
2)      Attribute
3)      Relationship

Q.1 (b)          Mention the 5 attributes types                                                                                  [2.5]

Attribute Type
1)      Simple
2)      Derived Attribute
3)      Complex Attribute
4)      Composite Attribute
5)      Multi-valued attribute

Q.1 (c)          Mention the 3 entity sets and explain them.                                                                 [3]
                       
An entity set is a set of entities of the same type.
Student: Set of BITS students
Faculty: Set of BITS faculties
Course: Set of BITS courses

Q.2 (a)          Create a relation(table) EMPLOYEE with Primary key emp_no                                 [4]
                          emp_no               fixed length  11 characters
                         emp_name           variable length of maximum 30 characters
                         designation           fixed length 10 characters’
                          salary                   number   (integer)
                          dept_no               number (integer)  with the value constraint 1 to 20


CREATE TABLE EMPLOYEE
(
            emp_no           CHAR(11),
emp_name       VARCHAR(30),
designation      CHAR(10),
            salary              INT,
                                    dept_no            INT CONSTRAINT CHK_DEPT CHECK (dept_no >= 1 AND dept_no <= 10),

CONSTRAINT PK_EMPLOYEE PRIMARY KEY CLUSTERED (emp_no)

)

Q.2 (b)          Create a relation(table) ORDER with primary key (order_no, cust_no, item_code) 
                   order_no               number(integer)
                   cust_no                  number(integer)
                   item_code            fixed length  10 characters
                   order_date            date
                  quantity                 number(integer)

by using the following relations(tables):
customer(cust_no, cust_name, address)
item(item_code, item_desc)                                                                                      [7]

CREATE TABLE ORDER
(
            order_no         INT,
cust_no            INT
REFERENCES customer.cust_no(cust_no),
item_code      CHAR(10)
REFERENCES item.item_code(item_code),
            order_date       date,
quantity           INT,

CONSTRAINT PK_ORDER PRIMARY KEY CLUSTERED (order_no, cust_no, item_code)
)



Q.3.        Write the SQL commands(statements) for the given relation below
         employee(emp_no, emp_name, designation, salary, dept_no)

(a)             Create a view CLERK_V containing the emp_name, designation and salary of employees for designation ‘CLERK’.

CREATE VIEW CLERK_V
AS
SELECT emp_name, salary
FROM employee
WHERE designation = ‘CLERK’





(b)             Update the salary of all employees whose designation is ‘SALESMAN’ working in the dept_no 20 with the salary 50000.                                                                   [4 + 3 = 7]


UPDATE employee
SET salary = 50000
WHERE designation = ‘SALESMAN’ AND dept_no = 20


Q.4.        student(st_id, name, birth_date, gender)                                                        
            course(course_no, title, units)
            st_course(st_id, course_no, sem, year, section_no, grade)

            Note: All the Computer Science courses start with CS,
                                  Electrical Engineering courses with  EE,
                                  Mechanical Engineering courses by ME

(a)              Write an SQL statement to list all the students with st_id, st_name whose name contains  KUMAR

SELECT st_id, name
FROM student
WHERE name LIKE ‘%KUMAR%’


(b)              Write an SQL statement to list all the Computer Science, Electrical Engineering  courses only with st_id, course_no, sem, year, section_no, grade, by arranging the course_no, section_no, st_id  in the ascending order.

SELECT s.st_id, sc.course_no, sc.sem, sc.year, sc.section_no, sc.grade
FROM student s INNER JOIN st_course sc ON s.st_id = sc.st_id
            INNER JOIN course c ON sc.course_no = c.course_no
WHERE c.title LIKE ‘CS%’ OR c.title LIKE ‘EE%’ 
ORDER BY course_no ASC, section_no ASC, st_id ASC


(c)              Write an SQL statement to list all students with st_id, name, course_no, who have taken more than 6 Computer Science courses.                                          [2 + 3 + 5 = 10]

SELECT s.st_id, s.name
FROM student s INNER JOIN st_course sc ON s.st_id = sc.st_id
INNER JOIN course c ON sc.course_no = c.course_no
            WHERE c.title LIKE ‘CS%’
GROUP BY s.st_id, s.name
HAVING COUNT(sc.course_no) >6

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