DBMS SQL

Database Management System
SQL
By:
Sharmila Chidaravalli
Asst. Prof.
Department of ISE
Global Academy of Technology

Dname location
civil Block B
CSE First Floor
ECE Second Floor
EEE Third Floor
ISE First Floor
Mech Block A
Department
USN name age semester Dname
CS025 Shilpa 22 7 CSE
EC006 Deepthi 20 3 ECE
IS001 Architha 20 5 ISE
IS025 Harsha 20 3 ISE
ME016 Akash 21 5 Mech
ME030 Nithin 21 5 Mech
ME045 Pramod 22 7 Mech
Student

Basic Retrieval Query : SELECT-FROM-WHERE SQL
select name
from student
where semester = 7;

select name ,Dname
from student
where semester = 5 and
Dname = ‘Mech’;

select name , Dname
from student
where semester = 5 or Dname = ‘Mech’;

select USN, name, Dname, location
from student, Department
where Dname = 'Mech' and Dname = Dname;
?

select USN, name, Dname, location
from student, Department
where Dname = 'Mech' and Dname = Dname;
ERROR 1052 (23000): Column 'Dname' in field list is ambiguous
?

Ambiguous Attribute Names, Aliasing, Renaming and tuple variable
select USN, name, student.Dname ,location
from student,Department
where student.Dname = 'Mech' and student.Dname = Department.Dname;

select S.name, S.semester, D.Dname, D.location
from student S, Department D
where S.Dname = D.Dname;

select USN from student;
+-------+
| USN |
+-------+
| CS025 |
| EC006 |
| IS001 |
| IS025 |
| ME016 |
| ME030 |
| ME045 |
+-------+
Unspecified WHERE Clause and Use of Asterisk

select * from student, Department;
+-------+----------+------+----------+-------+-------+--------------+
| USN | name | age | semester | Dname | Dname | location |
+-------+----------+------+----------+-------+-------+--------------+
| CS025 | Shilpa | 22 | 7 | CSE | civil | Block B |
| CS025 | Shilpa | 22 | 7 | CSE | CSE | First Floor |
| CS025 | Shilpa | 22 | 7 | CSE | ECE | Second Floor |
| CS025 | Shilpa | 22 | 7 | CSE | EEE | Third Floor |
| CS025 | Shilpa | 22 | 7 | CSE | ISE | First Floor |
| CS025 | Shilpa | 22 | 7 | CSE | Mech | Block A |
| EC006 | Deepthi | 20 | 3 | ECE | civil | Block B |
| EC006 | Deepthi | 20 | 3 | ECE | CSE | First Floor |
| EC006 | Deepthi | 20 | 3 | ECE | ECE | Second Floor |
| EC006 | Deepthi | 20 | 3 | ECE | EEE | Third Floor |
| EC006 | Deepthi | 20 | 3 | ECE | ISE | First Floor |
| EC006 | Deepthi | 20 | 3 | ECE | Mech | Block A |
| IS001 | Architha | 20 | 5 | ISE | civil | Block B |
| IS001 | Architha | 20 | 5 | ISE | CSE | First Floor |
| IS001 | Architha | 20 | 5 | ISE | ECE | Second Floor |
| IS001 | Architha | 20 | 5 | ISE | EEE | Third Floor |
| IS001 | Architha | 20 | 5 | ISE | ISE | First Floor |
| IS001 | Architha | 20 | 5 | ISE | Mech | Block A |
| IS025 | Harsha | 20 | 3 | ISE | civil | Block B |
| IS025 | Harsha | 20 | 3 | ISE | CSE | First Floor |
| IS025 | Harsha | 20 | 3 | ISE | ECE | Second Floor |
| IS025 | Harsha | 20 | 3 | ISE | EEE | Third Floor |
| IS025 | Harsha | 20 | 3 | ISE | ISE | First Floor |
| IS025 | Harsha | 20 | 3 | ISE | Mech | Block A |
| ME016 | Akash | 21 | 5 | Mech | civil | Block B |
| ME016 | Akash | 21 | 5 | Mech | CSE | First Floor |
| ME016 | Akash | 21 | 5 | Mech | ECE | Second Floor |
| ME016 | Akash | 21 | 5 | Mech | EEE | Third Floor |
| ME016 | Akash | 21 | 5 | Mech | ISE | First Floor |
| ME016 | Akash | 21 | 5 | Mech | Mech | Block A |
| ME030 | Nithin | 21 | 5 | Mech | civil | Block B |
| ME030 | Nithin | 21 | 5 | Mech | CSE | First Floor |
| ME030 | Nithin | 21 | 5 | Mech | ECE | Second Floor |
| ME030 | Nithin | 21 | 5 | Mech | EEE | Third Floor |
| ME030 | Nithin | 21 | 5 | Mech | ISE | First Floor |
| ME030 | Nithin | 21 | 5 | Mech | Mech | Block A |
| ME045 | Pramod | 22 | 7 | Mech | civil | Block B |
| ME045 | Pramod | 22 | 7 | Mech | CSE | First Floor |
| ME045 | Pramod | 22 | 7 | Mech | ECE | Second Floor |
| ME045 | Pramod | 22 | 7 | Mech | EEE | Third Floor |
| ME045 | Pramod | 22 | 7 | Mech | ISE | First Floor |
| ME045 | Pramod | 22 | 7 | Mech | Mech | Block A |
+-------+----------+------+----------+-------+-------+--------------+

Substring Pattern matching and Arthimetic Operators
select USN, name, Dname from student where name like '%a';

Substring Pattern matching and Arthimetic Operators
select USN, name, Dname from student where name like '%a';
Output:
| USN | name | Dname |
+-------+----------+-------+
| CS025 | Shilpa | CSE |
| IS001 | Architha | ISE |
| IS025 | Harsha | ISE |
+-------+----------+-------+

select USN from student where USN like '_S_ _ _';

select USN from student where USN like '_S_ _ _';
Output:
| USN |
+-------+
| CS025 |
| IS001 |
| IS025 |
+-------+

select name from student where name like '____h%';

select USN,name,semester+1 as New_sem from student;

select USN, name, semester+1 as New_sem
from student;
Output:
| USN | name | New_sem |
+-------+----------+---------+
| CS025 | Shilpa | 8 |
| EC006 | Deepthi | 4 |
| IS001 | Architha | 6 |
| IS025 | Harsha | 4 |
| ME016 | Akash | 6 |
| ME030 | Nithin | 6 |
| ME045 | Pramod | 8 |
+-------+----------+---------+

select USN,name,semester+1 as New_sem, location as department_location
where S.Dname = D.Dname and D.location = 'Block A';

select USN,name,semester+1 as New_sem, location as department_location
where S.Dname = D.Dname and D.location = 'Block A';
+-------+--------+---------+---------------------+
| USN | name | New_sem | department_location |
+-------+--------+---------+---------------------+
| ME016 | Akash | 6 | Block A |
| ME030 | Nithin | 6 | Block A |
| ME045 | Pramod | 8 | Block A |
+-------+--------+---------+---------------------+

Admin_id Name Salary
AD001 Kiran 10000
AD004 Mary 20000
AD010 Sandesh 40000
AD025 Krishna 25000
AD045 Sahana 20000
Admin_Staff
FID FNAme Salary DName
AD010 Sandesh 40000 EEE
CSE10 Kavitha 50000 CSE
CSE35 Priyanka 50000 CSE
ECE05 Ramya 45000 ECE
ECE25 Bharathi 50000 ECE
ISE25 Ashwini 45000 ISE
Faculty

SELECT DISTINCT Salary
FROM Faculty;
Table as Sets

SELECT DISTINCT Salary
FROM Faculty;
Output:
| Salary |
+--------+
| 40000 |
| 50000 |
| 45000 |
+--------+

SELECT ALL Salary
FROM Faculty;
Faculty

Faculty
Output:
| Salary |
+--------+
| 40000 |
| 50000 |
| 50000 |
| 45000 |
| 50000 |
| 45000 |
+--------+
SELECT ALL Salary
FROM Faculty;

(select distinct Salary from Faculty)
UNION
(select distinct Salary from Admin_Staff);

(select distinct Salary from Faculty)
UNION
(select distinct Salary from Admin_Staff);
+--------+
| Salary |
+--------+
| 40000 |
| 50000 |
| 45000 |
| 10000 |
| 20000 |
| 25000 |
+--------+

Output :
| Salary |
+--------+
| 40000 |
| 50000 |
| 45000 |
| 10000 |
| 20000 |
| 25000 |
+--------+
(select Salary from Faculty)
UNION
(select Salary from Admin_Staff);

UNION ALL

UNION ALL
Output :
| Salary |
+--------+
| 40000 |
| 50000 |
| 50000 |
| 45000 |
| 50000 |
| 45000 |
| 10000 |
| 20000 |
| 40000 |
| 25000 |
| 20000 |
+--------+

(select Fname from Faculty)
UNION
(select Name from Admin_Staff);

(select Name,Salary from Admin_Staff)
UNION
(select Fname,Salary from Faculty);

(select Name,Salary from Admin_Staff)
UNION
(select Fname,Salary from Faculty);
+----------+--------+
| Name | Salary |
+----------+--------+
| Kiran | 10000 |
| Mary | 20000 |
| Sandesh | 40000 |
| Krishna | 25000 |
| Sahana | 20000 |
| Kavitha | 50000 |
| Priyanka | 50000 |
| Ramya | 45000 |
| Bharathi | 50000 |
| Ashwini | 45000 |
+----------+--------+

select distinct Salary from Faculty inner join Admin_Staff
using(Salary);
+--------+
| Salary |
+--------+
| 40000 |
+--------+

select distinct Salary from Faculty left join Admin_Staff
using(Salary);
+--------+
| Salary |
+--------+
| 40000 |
| 50000 |
| 45000 |
+--------+

select Salary from Faculty left join Admin_Staff using(Salary);
+--------+
| Salary |
+--------+
| 40000 |
| 50000 |
| 50000 |
| 45000 |
| 50000 |
| 45000 |
+--------+

select Fname,Salary from Faculty order by Fname;
+----------+--------+
| Fname | Salary |
+----------+--------+
| Ashwini | 45000 |
| Kavitha | 50000 |
| Ramya | 45000 |
| Sandesh | 40000 |
+----------+--------+

select Fname,Salary from Faculty order by Salary ASC;
+----------+--------+
| Fname | Salary |
+----------+--------+
| Sandesh | 40000 |
| Ramya | 45000 |
| Ashwini | 45000 |
| Kavitha | 50000 |
+----------+--------+

Aggregate Functions in SQL
Select COUNT(FName) from Faculty;
+--------------+
| COUNT(FName) |
+--------------+
| 6 |
+--------------+

Select COUNT(FName) from Faculty where Dname='ECE';
+--------------+
| COUNT(FName) |
+--------------+
| 2 |
+--------------+

select SUM(Salary),MAX(Salary),MIN(Salary),AVG(Salary)
-> from Faculty;
+-------------+-------------+-------------+-------------+
| SUM(Salary) | MAX(Salary) | MIN(Salary) | AVG(Salary) |
+-------------+-------------+-------------+-------------+
| 280000 | 50000 | 40000 | 46666.6667 |
+-------------+-------------+-------------+-------------+

select SUM(Salary) AS SUM,MAX(Salary)AS
MAX_SAL,MIN(Salary)AS MIN_SAL,AVG(Salary) AS AVG_SAL
from faculty;
+--------+---------+---------+------------+
| SUM | MAX_SAL | MIN_SAL | AVG_SAL |
+--------+---------+---------+------------+
| 280000 | 50000 | 40000 | 46666.6667 |
+--------+---------+---------+------------+

select SUM(Salary) AS SUM,MAX(Salary)AS
MAX_SAL,MIN(Salary)AS MIN_SAL,AVG(Salary) AS AVG_SAL
from faculty where DName = 'CSE'
-> ;
+--------+---------+---------+------------+
| SUM | MAX_SAL | MIN_SAL | AVG_SAL |
+--------+---------+---------+------------+
| 100000 | 50000 | 50000 | 50000.0000 |
+--------+---------+---------+------------+

select Count(*) from Faculty;
+----------+
| Count(*) |
+----------+
| 6 |
+----------+

select Count(*) as NO_of_Faculties from Faculty;
+-----------------+
| NO_of_Faculties |
+-----------------+
| 6 |
+-----------------+

select Count(*) as No_of_Faculties,SUM(Salary) as SUM from
Faculty;
+-----------------+--------+
| No_of_Faculties | SUM |
+-----------------+--------+
| 6 | 280000 |
+-----------------+--------+

select Fname
from faculty
where(Salary between 35000 and 45000);

select Fname
from faculty
where(Salary not between 35000 and 45000);

Group By & Having Clauses in SQL
select dname ,Count(Dname) as NO_of_Students
from student
group by Dname;

from student group by Dname;
| dname | NO_of_Students |
+-------+----------------+
| CSE | 3 |
| ECE | 2 |
| ISE | 2 |
| Mech | 3 |
+-------+----------------+

from student
where semester = 7
group by Dname;

from student
where semester = 7
group by Dname;
Output:
+-------+----------------+
| CSE | 2 |
| Mech | 1 |
+-------+----------------+

select dname , semester, Count(*) as NO_of_Students
from student
group by Dname, semester;

select dname , semester, Count(*) as NO_of_Students
from student
group by Dname, semester;
Output:
| dname | semester | NO_of_Students |
+-------+----------+----------------+
| CSE | 5 | 1 |
| CSE | 7 | 2 |
| ECE | 3 | 1 |
| ECE | 5 | 1 |
| ISE | 3 | 1 |
| ISE | 5 | 1 |
| Mech | 5 | 2 |
| Mech | 7 | 1 |
+-------+----------+----------------+

select dname ,semester, Count(*) as NO_of_Students
from student
group by semester;

select dname ,semester, Count(*) as NO_of_Students
from student
group by semester;
Output:
| dname | semester | NO_of_Students |
+-------+----------+----------------+
| ECE | 3 | 2 |
| CSE | 5 | 5 |
| CSE | 7 | 3 |
+-------+----------+----------------+

from student
group by Dname
having count(dname)>2;

from student
group by Dname
having count(dname)>2;
Output:
+-------+----------------+
| CSE | 3 |
| Mech | 3 |
+-------+----------------+

from student
group by Dname
having count(dname)>2
order by Dname DESC;

from student
group by Dname
having count(dname)>2
order by Dname DESC;
Output:
+-------+----------------+
| Mech | 3 |
| CSE | 3 |
+-------+----------------+

select dname ,Count(*) as NO_of_Students
from student
where semester = 7
group by Dname
having count(*)>1;

select dname ,Count(*) as NO_of_Students
from student
where semester = 7
group by Dname
having count(*)>1;
Output :
+-------+----------------+
| CSE | 2 |
+-------+----------------+

select DName,Count(*) as No_of_Faculties, AVG(salary)
from faculty
group by DName;

select DName,Count(*) as No_of_Faculties,AVG(salary)
from faculty
group by DName;
Output:
| DName | No_of_Faculties | AVG(salary) |
+-------+-----------------+-------------+
| CSE | 2 | 50000.0000 |
| ECE | 2 | 47500.0000 |
| EEE | 1 | 40000.0000 |
| ISE | 1 | 45000.0000 |
+-------+-----------------+-------------+

select DName, Count(*) as No_of_Faculties, AVG(salary)
from faculty
group by Dname
having avg(salary)< 50000;

select DName,Count(*) as No_of_Faculties,AVG(salary)
from faculty
group by DName
having avg(salary)< 50000;
Output:
+-------+-----------------+-------------+
| ECE | 2 | 47500.0000 |
| EEE | 1 | 40000.0000 |
| ISE | 1 | 45000.0000 |
+-------+-----------------+-------------+

Select DName,Count(*) as No_of_Faculties,AVG(salary)
from faculty
where salary >= 45000
group by DName
having avg(salary) < 50000;
Output:
+-------+-----------------+-------------+
| ECE | 2 | 47500.0000 |
| ISE | 1 | 45000.0000 |
+-------+-----------------+-------------+

create view CSE as select USN,name
from Student
where Dname ='CSE' or Dname = 'ISE';
Views (Virtual Tables) in SQL

create view CSE as
select USN,name
from Student
where Dname ='CSE' or Dname = 'ISE';
mysql> select * from CSE;
USN | name |
+-------+-----------+
| CS006 | Sharath |
| CS025 | Shilpa |
| CS045 | Rakshitha |
| IS001 | Architha |
| IS025 | Harsha |
+-------+-----------+

create view Sem_5 as
select USN ,name
from student
where semester = 5;
select * from Sem_5;
Output:
| USN | name |
+-------+-----------+
| EC020 | Sushma |
| ME016 | Akash |
| ME030 | Nithin |
+-------+-----------+

select C.USN,C.name
from CSE C,Sem_5 S
where C.USN = S.USN;
Output:
| USN | name |
+-------+-----------+
+-------+-----------+

select B.bid from boats B where B.color ='red';
+-----+
| bid |
+-----+
| 102 |
| 104 |
+-----+
select R.sid from reserves R where R.bid IN(102,104);
+-----+
| sid |
+-----+
| 22 |
| 31 |
| 64 |
| 22 |
+-----+
select S.sname from sailors S where S.sid NOT IN(22,31,64,22);
+---------+
| sname |
+---------+
| Brutus |
| Andy |
| Rusty |
| Zorba |
| Horatio |
| Art |
| Bob |
+---------+

Nested Queries – Set Comparison Operators
Find sailors whose rating is better than some sailor called Horatio
-----+
| sid |
+-----+
| 31 |
| 32 |
| 58 |
| 71 |
| 74 |
+-----+

Find sailors whose rating is better than every sailor called Horatio
select S.sid
from sailors S
where S.rating > ALL(select S2.rating
from sailors S2
where S2.sname = 'Horatio');
OUTPUT:
| sid |
+-----+
| 58 |
| 71 |
+-----+

Find the sailors with the highest rating
select S.sid
from sailors S
where S.rating >= ALL(select S2.rating
from sailors S2 );
+-----+
| sid |
+-----+
| 58 |
| 71 |
+-----+

Find the names of the sailors who have reserved all boats

Find the names of the sailors who have reserved all boats
select S.sname
from sailors S
where NOT EXISTS (select B.bid
from boats B
where NOT Exists (select R.bid
from reserves R
where R.bid=B.bid and R.sid=S.sid));
--------+
| sname |
+--------+
| Dustin |
+--------+

Find the names of sailors who are older than the oldest
sailor with a rating of 10.
SELECT S.sname
FROM Sailors S
WHERE S.age > ( SELECT MAX( S2.age )
FROM Sailors S2
WHERE S2.rating = 10 );
| sname |
+--------+
| Dustin |
| Lubber |
| Bob |
+--------+

CountryId CountryName
1 INDIA
2 Nepal
3 Srilanka
Country
StateID CountryId StateName
1 1 Karnataka
2 1 Mumbai
3 2 Katmandu
4 Null California
State
Joins in SQL
INNER JOIN
LEFT OUTER JOIN
RIGHT OUTER JOIN

Joins in SQL
select *
from Country C INNER JOIN State S
on C.CountryId = S.CountryId;

Joins in SQL
select *
OUTPUT:
| CountryId | CountryName | StateID | CountryId | StateName |
+-----------+-------------+---------+-----------+-----------+
| 1 | INDIA | 1 | 1 | Karnataka |
| 1 | INDIA | 2 | 1 | Mumbai |
| 2 | Nepal | 3 | 2 | Katmandu |
+-----------+-------------+---------+-----------+-----------+

Joins in SQL
select *
from Country C LEFT OUTER JOIN State S

Joins in SQL
select *
OUTPUT:
+-----------+-------------+---------+-----------+-----------+
| 1 | INDIA | 2 | 1 | Mumbai |
| 3 | Srilanka | NULL | NULL | NULL |
+-----------+-------------+---------+-----------+-----------+

Joins in SQL
select *
from Country C RIGHT OUTER JOIN State S

Joins in SQL
select *
on C.CountryId = S.CountryId
UNION
select *

In MySQL, a trigger is a set of SQL statements that is invoked automatically when a change is
made to the data on the associated table.
A trigger can be defined to be invoked either before or after the data is changed by INSERT,
UPDATE or DELETE statement.
BEFORE INSERT - activated before data is inserted into the table.
AFTER INSERT - activated after data is inserted into the table.
BEFORE UPDATE - activated before data in the table is updated.
AFTER UPDATE - activated after data in the table is updated.
BEFORE DELETE - activated before data is removed from the table.
AFTER DELETE - activated after data is removed from the table.
TRIGGERS

When you use a statement that does not use INSERT, DELETE or UPDATE
statement to change data in a table, the triggers associated with the table are not
invoked.
MySQL Trigger Syntax
CREATE TRIGGER trigger_name trigger_time trigger_event
ON table_name
FOR EACH ROW
BEGIN
.........
END;

Trigger activation time can be BEFORE or AFTER. You must specify the activation time when you define a
trigger. You use the BEFORE keyword if you want to process action prior to the change is made on the table
and AFTER if you need to process action after the change is made.
The trigger event can be INSERT, UPDATE or DELETE. This event causes the trigger to be invoked. A
trigger only can be invoked by one event. To define a trigger that is invoked by multiple events, you have to
define multiple triggers, one for each event.
A trigger must be associated with a specific table. Without a table trigger would not exist therefore you have
to specify the table name after the ON keyword.
You place the SQL statements between BEGIN and END block. This is where you define the logic for the
trigger.

itemid iDesc qoh price category
8 perfume 59 380 Cosemetics
10 kb 59 180 toys
Item
sid itemid qtysold price total
101 8 10 90 0
Sales_new
create trigger t2
before insert
on Sales_new
for each row
begin
set new.total = new.qtysold * new.price;
end;

10 kb 59 180 toys
Item
101 8 10 90 0
Sales_new
insert into Sales_new values(102,10,10,180,0);

10 kb 59 180 toys
Item
101 8 10 90 0
102 10 10 180 1800
Sales_new
select * from Sales_new;

10 kb 59 180 toys
Item
101 8 10 90 0
102 8 10 90 900
create trigger t1
before insert
on Sales
for each row
begin
set new.total=new.qtysold*new.price;
update item
set qoh=qoh-new.qtysold
where item.itemid=new.itemid;
end;
Sales

10 kb 59 180 toys
Item
101 8 10 90 0
102 8 10 90 900
Sales
insert into Sales values(103,10,15,180,0);

10 kb 44 180 toys
Item
101 8 10 90 0
102 8 10 90 900
103 10 15 180 2700
Sales
select * from Sales;
select * from Item;

Assignment
Consider the following relations:
Student(snum: integer, sname: string, major: string, level: string, age: integer)
Class(name: string, meets_at: time, room: string, fid: integer)
Enrolled(snum: integer, cname: string)
Faculty(fid: integer, fnarne: string, deptid: integer)
The meaning of these relations is straightforward; for example, Enrolled has one record per student-class pair
such that the student is enrolled in the class.
Write the following queries in SQL. No duplicates should be printed in any of the answers.
1. Find the names of all Juniors (level = JR) who are enrolled in a class taught by Prof.John Teach.
2. Find the age of the oldest student who is either a History major or enrolled in a course taught by 1. Teach.
3. Find the names of all classes that either meet in room R128 or have five or more students enrolled.
4. Find the names of all students who are enrolled in two classes that meet at the same time.
Submission Date : 10-9-2019
Time : 1.00 PM

5. Find the names of faculty members who teach in every room in which some class is taught.
6. Find the names of faculty members for whom the combined enrollment of the courses that they teach is
less than five.
7. Print the level and the average age of students for that level, for each level.
8. Print the level and the average age of students for that level, for all levels except JR.
9. For each faculty member that has taught classes only in room R128, print the faculty member's name and
the total number of classes she or he has taught.
10. Find the names of students enrolled in the maximum number of classes.
11. Find the names of students not enrolled in any class.
12. For each age value that appears in Students, find the level value that appears most often. For example, if
there are more FR level students aged 18 than SR, JR, or SO students aged 18, you should print the pair (18,
FR).

DBMS SQL

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