MySQL Aggregate Function Exercises: Get the number of employees with the same job

Last update on February 02 2026 11:01:42 (UTC/GMT +8 hours)

MySQL Aggregate Function: Exercise-7 with Solution

Write a query to get the number of employees with the same job.

Sample table: employees

+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
| EMPLOYEE_ID | FIRST_NAME  | LAST_NAME   | EMAIL    | PHONE_NUMBER       | HIRE_DATE  | JOB_ID     | SALARY   | COMMISSION_PCT | MANAGER_ID | DEPARTMENT_ID |
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
|         100 | Steven      | King        | SKING    | 515.123.4567       | 1987-06-17 | AD_PRES    | 24000.00 |           0.00 |          0 |   		  90 |
|         101 | Neena       | Kochhar     | NKOCHHAR | 515.123.4568       | 1987-06-18 | AD_VP      | 17000.00 |           0.00 |        100 |            90 |
|         102 | Lex         | De Haan     | LDEHAAN  | 515.123.4569       | 1987-06-19 | AD_VP      | 17000.00 |           0.00 |        100 |            90 |
|         103 | Alexander   | Hunold      | AHUNOLD  | 590.423.4567       | 1987-06-20 | IT_PROG    |  9000.00 |           0.00 |        102 |            60 |
|         104 | Bruce       | Ernst       | BERNST   | 590.423.4568       | 1987-06-21 | IT_PROG    |  6000.00 |           0.00 |        103 |            60 |
|         105 | David       | Austin      | DAUSTIN  | 590.423.4569       | 1987-06-22 | IT_PROG    |  4800.00 |           0.00 |        103 |            60 |
|         106 | Valli       | Pataballa   | VPATABAL | 590.423.4560       | 1987-06-23 | IT_PROG    |  4800.00 |           0.00 |        103 |            60 |
|         107 | Diana       | Lorentz     | DLORENTZ | 590.423.5567       | 1987-06-24 | IT_PROG    |  4200.00 |           0.00 |        103 |            60 |
|         108 | Nancy       | Greenberg   | NGREENBE | 515.124.4569       | 1987-06-25 | FI_MGR     | 12000.00 |           0.00 |        101 |           100 |
.........
|         206 | William     | Gietz       | WGIETZ   | 515.123.8181       | 1987-10-01 | AC_ACCOUNT |  8300.00 |           0.00 |        205 |           110 |
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+

View the table

Code:

-- Counting the number of employees for each job ID
SELECT job_id, COUNT(*) 
-- Selecting data from the employees table
FROM employees 
-- Grouping the result set by the job_id column
GROUP BY job_id;

Explanation:

This SQL query counts the number of employees for each unique job ID.
The SELECT statement retrieves the job_id column and counts the number of rows for each job ID using the COUNT(*) function.
The FROM clause specifies the employees table from which the data is being selected.
The GROUP BY clause groups the result set by the job_id column, ensuring that the count is calculated for each distinct job ID.

Relational Algebra Expression:

Relational Algebra Tree:

Pictorial Presentation of the above query

Go to:

PREV :Write a query to get the highest, lowest, sum, and average salary of all employees.
NEXT :Write a query to get the difference between the highest and lowest salaries.

MySQL Code Editor:

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