PostgreSQL AVG function

AVG function

The AVG function determines the average of all selected values of a column.

Syntax:

AVG (* | [DISTINCT] ALL | column_name)

Parameters

PostgreSQL AVG function example

The sample table

If we want to find the average salary for all employees in the employee table, the following SQL can be used.

SQL

Code:

SELECT AVG(salary) 
FROM employee;

Output:

Pictorial Presentation of PostgreSQL AVG() function

PostgreSQL AVG as a level

If we want to get the average salary for all employees and show the result against 'Average Salary' head in the employee table, the following SQL can be used.

SQL

Code:

SELECT AVG(salary) "Average Salary" 
FROM employee;

OR

Code:

SELECT AVG(DISTINCT salary) "Average Salary" 
FROM employee;

OR

Code:

SELECT AVG(ALL salary) "Average Salary" 
FROM employee;

Output:

PostgreSQL AVG two columns

If we want to get the average salary and deduction from employee table, the following SQL can be used.

SQL

Code:

SELECT AVG(salary) "Average Salary" ,
AVG(deduction) "Average Deduction"
FROM employee;

Output:

PostgreSQL AVG with MAX, MIN, COUNT and SUM

If we want to get the average salary, maximum salary, minimum salary, total employee and total salary from employee table, the following SQL can be used.

SQL

Code:

SELECT AVG(salary) "Average Salary" ,
MAX(salary) "Maximum Salary",
MIN(salary) "Minimum Salary",
COUNT(*) "Total Employee",
SUM(salary) "Total Salary"
FROM employee;

Output:

PostgreSQL AVG WHERE clause

If we want to get the average salary and deduction from employee table who belongs to the designation 'SALESMAN', the following SQL can be used.

SQL

Code:

SELECT AVG(salary) "Average Salary" ,
AVG(deduction) "Average Deduction"
FROM employee
WHERE designame='SALESMAN';

Output:

Pictorial Presentation of PostgreSQL AVG with WHERE clause

PostgreSQL AVG with GROUP BY

Sample table: employees

The following statement will return the designation and average salary against each designation from the employees table.

SQL

Code:

SELECT job_id, ROUND(AVG(salary),2) AS "Average Salary rounded" 
FROM employees 
GROUP BY job_id;

Sample Output:

   job_id   | Average Salary
------------+----------------
 AC_ACCOUNT |        8300.00
 ST_MAN     |        7280.00
 IT_PROG    |        5760.00
 SA_MAN     |       12200.00
 AD_PRES    |       24000.00
 AC_MGR     |       12000.00
 FI_MGR     |       12000.00
 AD_ASST    |        4400.00
 MK_MAN     |       13000.00
 PU_CLERK   |        2780.00
 HR_REP     |        6500.00
 PR_REP     |       10000.00
 FI_ACCOUNT |        7920.00
 SH_CLERK   |        3215.00
 AD_VP      |       17000.00
 SA_REP     |        8350.00
 ST_CLERK   |        2785.00
 MK_REP     |        6000.00
 PU_MAN     |       11000.00
(19 rows)

Pictorial Presentation of PostgreSQL AVG with GROUP BY

PostgreSQL AVG with WHERE CLAUSE and GROUP BY

Sample table: employees

The following statement will return the designation wise average salary for that designation which salary does not exceed 12000 and above, from employees table.

SQL

Code:

SELECT job_id,ROUND(AVG(salary),2) AS "Avereage Salary" 
FROM employees 
WHERE salary<12000 
GROUP BY job_id;

Sample Output:

   job_id   | Avereage Salary
------------+-----------------
 AC_ACCOUNT |         8300.00
 ST_MAN     |         7280.00
 IT_PROG    |         5760.00
 SA_MAN     |        10750.00
 AD_ASST    |         4400.00
 PU_CLERK   |         2780.00
 HR_REP     |         6500.00
 PR_REP     |        10000.00
 FI_ACCOUNT |         7920.00
 SH_CLERK   |         3215.00
 SA_REP     |         8350.00
 ST_CLERK   |         2785.00
 MK_REP     |         6000.00
 PU_MAN     |        11000.00
(14 rows)

PostgreSQL AVG with HAVING and GROUP BY

Sample table : employees

The following sql statement will return, designation wise total employees and the average salary for that designation which salary does not exceed 12000 and above and a minimum of five employees have this designation.

SQL

Code:

SELECT job_id,COUNT(*) AS "Number of Employees", 
ROUND(AVG(salary),2) AS "Average Salary" 
FROM employees 
WHERE salary<12000 
GROUP BY job_id 
HAVING COUNT(*)>=5;

Sample Output:

   job_id   | Number of Employees | Average Salary
------------+---------------------+----------------
 ST_MAN     |                   5 |        7280.00
 IT_PROG    |                   5 |        5760.00
 PU_CLERK   |                   5 |        2780.00
 FI_ACCOUNT |                   5 |        7920.00
 SH_CLERK   |                  20 |        3215.00
 SA_REP     |                  30 |        8350.00
 ST_CLERK   |                  20 |        2785.00
(7 rows)			  

Pictorial Presentation of PostgreSQL AVG with HAVING

PostgreSQL AVG with GROUP BY and ORDER BY

Sample table: employees

The following statement will return, the designation wise total employees and average salary, arranged in descending order according to the total number of employees, for that designation which salary does not exceed 12000 and above and a minimum of five employees has this designation.

SQL

Code:

SELECT job_id,COUNT(*) AS "Number of employees" ,
ROUND(AVG(salary),2) AS "Average Salary" 
FROM employees 
WHERE salary<12000 
GROUP BY job_id 
HAVING COUNT(*)>=5 
ORDER BY COUNT(*) DESC;

Sample Output:

   job_id   | Number of employees | Average Salary
------------+---------------------+----------------
 SA_REP     |                  30 |        8350.00
 SH_CLERK   |                  20 |        3215.00
 ST_CLERK   |                  20 |        2785.00
 ST_MAN     |                   5 |        7280.00
 FI_ACCOUNT |                   5 |        7920.00
 IT_PROG    |                   5 |        5760.00
 PU_CLERK   |                   5 |        2780.00
(7 rows)

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