SQL and E R diagram

7/11/2016 Prepared By- Shimul & Hirok,CSE,MBSTU 1
WELCOME
TO OUR
PRESENTATION

SQL and E-R diagram related
problems and its solution.
7/11/2016 2Prepared By- Shimul & Hirok,CSE,MBSTU
Presentation on……
Presented By
• NAME : ID:
• MD.MAHBUBUR RAHMAN CE-12038
• HIROK BISWAS CE-12032

We are going to talking about
 SQL expressions:
 sample bank database(3.19)
 relational schema(3.21,3.22)
E-R diagram for a hospital(6.15)
UML equivalents of the E-R diagrams(6.28)
Conclution

Problem 1 :
Using the relations of our sample bank database, write SQL expressions to
define the following views:
a. A view containing the account numbers and customer names (but not the
balances) for all accounts at the Deer Park branch.
b. A view containing the names and addresses of all customers who have an
account with the bank, but do not have a loan.
c. A view containing the name and average account balance of every customer
of the Rock Ridge branch.

Solution of Problem 1 (a)
A view containing the account numbers and customer names (but not
the balances) for all accounts at the Deer Park branch.
Table: accountTable: depositor

Solution of Problem 1 (a)( Cont…)
• Query:
select depositor.customer_name,depositor.account_number
from depositor,account
where depositor.account_number =account.account_number
and account.branch_name='Deer Park';
• Table view:

Solution of Problem 1 (b)
A view containing the names and addresses of all customers who have
an account with the bank, but do not have a borrower.
Table: customer Table: depositor Table: borrower

Solution of Problem 1 (b)(Cont…)
select c.customer_name, c.customer_street,
c.customer_city
from customer c, depositor d
where c.customer_name not in (select
customer_name
from borrower)
and c.customer_name = d.customer_name;
• Query:
• Table view:

Solution of Problem 1 (c)
. A view containing the name and average account balance of every
customer of the 'sidney' branch.
Table: account Table: depositor

Solution of Problem 1 (c) (Cont…)
select customer_name, avg(a.balance)
from account a, depositor d
where a.account_number = d.account_number
and branch_name='sidney'
group by customer_name;
• Query:
• Table view:

Problem 1 (EXP-02) :
create procedure get_emp(employee_id in varchar(10))
return char
BEGIN
select employee.employee_id,department.department_name,salary.salary_scale,salary.salary_amount from department
inner join employee_department on department.department_id=employee_department.department_id
inner join employee on employee.employee_id=employee_department.employee_id
inner join employee_salary on employee.employee_id=employee_salary.employee_id
inner join salary on salary.salary_scale=employee_salary.salary_scale
where employee.employee_id=employee_id;
return(employee_id);
END;
Exec get_emp('emp02');
Solution of Problem 1 (Exp-2):
Create a procedure that will take Employee_ID it’s parameter return
Employee_ID ,Department_Name, Salary_Scale and Salary_Amount.

Problem 2 :
Consider the following relational schema
employee(emp_no , name, office, age)
books(isbn , title, authors, publisher)
loan(emp_no , isbn, date)
Write the following queries in SQL.
a. Print the names of employees who have borrowed any book published by
McGraw-Hill.
b. Print the names of employees who have borrowed all books published by
McGraw-Hill.
c. For each publisher, print the names of employees who have borrowed more than five
books of that publisher.

Print the names of employees who have borrowed any book published
by McGraw-Hill.
Table: employee
Table: books
Table: loan

Solution of Problem 2 (a) (cont…)
• Query:
select distinct name as employee_name from employee
inner join loan on employee.emp_no=loan.emp_no
inner join books on loan.isbn=books.isbn
where books.publisher='Mcgraw-hill';
• Table view:

Print the names of employees who have borrowed all book published
by McGraw-Hill.
Table: employee
Table: books
Table: loan
result

Solution of Problem 2 (b) (cont…)
• Query:
select distinct name as employee_name from employee
inner join loan on employee.emp_no=loan.emp_no
inner join books on loan.isbn=books.isbn
where (select count(loan.isbn) from loan where
loan.emp_no=employee.emp_no)=(select count(books.isbn) from books
where books.publisher='Mcgraw-hill');
• Table view:

Solution of Problem 2 (c)
For each publisher, print the names of employees who have borrowed
more than five books of that publisher.
Table: employee
Table: books
Table: loan
result
More than
5
Books from
Mcgraw-hill
(only)

Solution of Problem 2 (c) (cont…)
• Query:
select name from employee,loan,books
where employee.emp_no=loan.emp_no and
books.isbn=loan.isbn
group by employee.emp_no,name,books.publisher having
count(loan.isbn) >=5;
• Table view:

Problem 3 :
Consider the relational schema
student(student_id , student_name)
registerd(student_id , course_id)
Write an SQL query to list the student-id and name of each
student along with the total number of courses that the student is
registered for. Students who are not registerd for any course must
also be listed, with the number of registerd
courses shown as 0.

Solution of Problem 3
Table: student
Table: registerd
a is registerd by
3
course
b is registerd by
1
course
C is not
registerd
(0)

Solution of Problem 3 (cont…)
• Query:
select student_id,student_name,(select count(registerd.course_id) from registerd
where student.student_id=registerd.student_id) as taken_course
from student;
• Table view:

Problem 4:
Consider the relational schema:
Department (Department _ID , Department _Name)
Employee (Employee_ID , Employee_Name,Employee_Age)
Employee _Department( Employee _ID , Department _ID )
Salary (Salary _Scale , Salary _Amount)
Employee_ Salary (Employee _ID , Salary _Scale)
a) Find the Department_names where the employee’s average salary is more
than 10000.
b) Find the employee_names who has paid in salary scale 1.
c) Update the employee ages by increasing 10 years whose ages are not equal
to average age.

Find the Department_names where the employee’s average salary is more
than 10000.
Table: department Table: employee_department
Table: employee_salary
Table: salary

Solution of Problem 4(a) (cont…)
• Query:
select department_name,avg(salary.salary_amount) from department
inner join employee_department on
department.department_id=employee_department.department_id
inner join employee_salary on
employee_department.employee_id=employee_salary.employee_id
having avg(salary.salary_amount)>10000
group by department.department_name;
• Table view:

Find the employee_names who has paid in salary scale 1.
Table: Employee Table: employee_salary
Table: salary
result

Solution of Problem 4(b) (cont…)
• Query:
select employee_name from employee inner join employee_salary
on employee_salary.employee_id=employee.employee_id
where salary.salary_scale=1;
• Table view:

Solution of Problem 4(c)
• Query:
update employee
set employee.employee_age=employee.employee_age+1
where employee.employee_age not in (select avg(employee.employee_age) from employee);
• Table view:
Before update:
After update:
Average
age

Problem 5
Construct an E-R diagram for a hospital with a
set of patients and a set of medical doctors.
Associate with each patient a log of the
various tests and examinations conducted.

Solution Of problem 5:
Fig: E-R diagram for a hospital system.

Problem 6
Draw the UML equivalents of the
E-R diagrams Of Figures
6.8c,6.9,6.17,6.72,And 6.20.

Needed Figure of Problem 6 And Its
Solution Given…………………….

Fig:6.8c Relationship(One –to-one)
Solution of Fig 6.8c:

Solution of Fig 6.9:

Solution of Fig 6.20(Cont….)

THANK
YOU

SQL and E R diagram

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