sql function

1. AGGREGATE FUNCTIONS
Prof. Sin-Min Lee
Surya Bhagvat
CS 157A – Fall 2005

2. Aggregate Functions
 What is an aggregate function?
An aggregate function summarizes the results
of an expression over a number of rows,
returning a single value. The general syntax for
most of the aggregate functions is as follows:
aggregate_function ([DISTINCT|ALL]
expression)

3. Commonly used Aggregate
functions
Some of the commonly used aggregate functions
are :
• SUM
• COUNT
• AVG
• MIN
• MAX

4. Examples
Consider the following Employee table:
EMPLOYEE ( EMP_ID, NAME, DEPT_NAME, SALARY)
CREATE TABLE EMPLOYEE
(
EMP_ID NUMBER,
NAME VARCHAR2(50),
DEPT_NAME VARCHAR2(50),
SALARY NUMBER
);

5. Employee Table (Contd….)
Run the following script to insert the records in the table
INSERT INTO EMPLOYEE VALUES (100,'ABC','ENG',50000);
INSERT INTO EMPLOYEE VALUES (101,'DEF','ENG',60000);
INSERT INTO EMPLOYEE VALUES (102,'GHI','PS',50000);
INSERT INTO EMPLOYEE VALUES (103,'JKL','PS',70000);
INSERT INTO EMPLOYEE VALUES (104,'MNO','SALES',75000);
INSERT INTO EMPLOYEE VALUES (105,'PQR','MKTG',70000);
INSERT INTO EMPLOYEE VALUES (106,‘STU','SALES',null);
COMMIT;

6. Select on Employee Table
After the insert when we query the Employee table we get the
following results:
Select * from Employee;

7. Performing SUM
Query 1: To find the sum of all salaries in the organization:
SELECT SUM(SALARY) FROM EMPLOYEE;
375000
Query 2: To find the sum of the salaries grouped by dept
SELECT SUM(SALARY) FROM EMPLOYEE GROUP BY
DEPT_NAME

8. SUM (Continued)
If we take a look at the previous query the information won’t
tell us what’s the sum for a particular department. So to include that
information we add DEPT_NAME in the SELECT
SELECT DEPT_NAME,SUM(SALARY) FROM EMPLOYEE
GROUP BY DEPT_NAME;

9. SUM (Continued…..)
The query in the previous slide lists the information for all the
departments. What if we want the information to be restricted only
for a particular department like Engg
Is this query correct?
SELECT DEPT_NAME,SUM(SALARY) FROM EMPLOYEE
GROUP BY
DEPT_NAME
WHERE DEPT_NAME = 'ENG';

10. SUM (Continued….)
No, the query would result in the sql error (in Oracle)
ORA-00933: SQL Command not properly ended
Remember : If we use the aggregate functions then you cannot use
the WHERE clause. In order to get the result what we need to use is
the HAVING clause. So the query would be
SELECT DEPT_NAME,SUM(SALARY) FROM EMPLOYEE
GROUP BY
DEPT_NAME
HAVING DEPT_NAME = 'ENG';

11. AVG Function
Query 1: If we want to calculate the AVG of all the salaries in
the organization the SQL would be
SELECT AVG(SALARY) FROM EMPLOYEE
62,500
Is this what we expect????
Employee table has 7 records and the salaries are
50,000+60,000+50,000+70,000+75,000+70,000+null/7 = 53571
But we obtained 62500 from the query? Why is this so????

12. AVG (Continued….)
Remember : COUNT(*) is the only function which won’t ignore
Nulls. Other functions like SUM,AVG,MIN,MAX they ignore
Nulls. What it means is in the previous query the salary value for
a particular employee was NULL. So the query
SELECT AVG(SALARY) FROM EMPLOYEE
would ignore nulls and the way the average is calculated then would
be
50,000+60,000+50,000+70,000+75,000+70,000/6 = 62500

13. AVG (Continued….)
From the information given in the previous slide what do you think
would be the output of the following query
Select COUNT(*),COUNT(SALARY) FROM EMPLOYEE;
It would be
COUNT(*) COUNT(SALARY)
7 6
Because COUNT(*) is not going to ignore the Nulls in the result
whereas COUNT(SALARY) is going to ignore the Nulls.

14. AVG (Continued…..)
SELECT student_name,avg(mark) FROM student,enrolment
WHERE student.student_id=enrolment.student_id;
Which one of the following is correct for the query?
(a) The query is not legal
(b) The query retrieves for each student enrolled,his/her name and their
average mark
(c) The query retrieves for each student enrolled,his/her name and the clas
average mark
(d) The query retrieves for each student enrolled,his/her name and the
mark in each subject
Is the answer (a) or (b)??????

15. AVG (Continued….)
If option 1 is not given then the correct answer would be option 2.
//Script begin
Drop table student;
Drop table enrolment;
create table Student
(student_name varchar2(100),
student_id varchar2(50)
);
create table enrolment
(student_id varchar2(50),
mark number);

16. AVG (Continued….)
//Script Continued
insert into student values ('A','1');
insert into student values ('B','2');
insert into student values ('C','3');
insert into enrolment values ('1',10);
insert into enrolment values ('1',20);
insert into enrolment values ('1',30);
insert into enrolment values ('2',40);
insert into enrolment values ('2',50);
insert into enrolment values ('2',60);
insert into enrolment values ('3',70);
insert into enrolment values ('3',60);
insert into enrolment values ('3',50);
commit;

17. AVG (Continued….)
If we try to execute the query given in the question
SELECT student_name,avg(mark) FROM student,enrolment
WHERE student.student_id=enrolment.student_id;
We would get the following error in Oracle
ORA-00937:not a single-group group function
Why is it so????

18. AVG (Continued….)
Remember : When we use any of the aggregate functions in SQL
all the columns listed in the SELECT need to be part of the
GROUP BY Clause. In the previous SQL
SELECT student_name,avg(mark) FROM student,enrolment
WHERE student.student_id=enrolment.student_id;
student_name, avg(mark) are the columns included in the select.
avg is the aggregate function. So if we leave that one out then
the column which needs to part of the group by clause would be
student_name.

19. AVG (Final SQL)
The final SQL then would be
SELECT student_name,avg(mark)
FROM student,enrolment
WHERE student.student_id=enrolment.student_id
group by student_name;
Which would give out the desired output

20. Using MIN AND MAX
Query 1: To find the minimum salary within a particular
department
SELECT MIN(SALARY),NAME FROM EMPLOYEE
GROUP BY NAME;
Query 2: To find the maximum salary within a particular
department
SELECT MAX(SALARY),NAME FROM EMPLOYEE
GROUP BY NAME;