SQL Exercise: Departments more than 10 employees who get commissions

Last update on January 08 2026 12:47:52 (UTC/GMT +8 hours)

31. From the following table, write a SQL query to find the departments where more than ten employees receive commissions. Return department id.

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       | 2003-06-17 | AD_PRES    | 24000.00 |           0.00 |          0 |            90 |
|         101 | Neena       | Kochhar     | NKOCHHAR | 515.123.4568       | 2005-09-21 | AD_VP      | 17000.00 |           0.00 |        100 |            90 |
|         102 | Lex         | De Haan     | LDEHAAN  | 515.123.4569       | 2001-01-13 | AD_VP      | 17000.00 |           0.00 |        100 |            90 |
|         103 | Alexander   | Hunold      | AHUNOLD  | 590.423.4567       | 2006-01-03 | IT_PROG    |  9000.00 |           0.00 |        102 |            60 |
|         104 | Bruce       | Ernst       | BERNST   | 590.423.4568       | 2007-05-21 | IT_PROG    |  6000.00 |           0.00 |        103 |            60 |
|         105 | David       | Austin      | DAUSTIN  | 590.423.4569       | 2005-06-25 | IT_PROG    |  4800.00 |           0.00 |        103 |            60 |
|         106 | Valli       | Pataballa   | VPATABAL | 590.423.4560       | 2006-02-05 | IT_PROG    |  4800.00 |           0.00 |        103 |            60 |
|         107 | Diana       | Lorentz     | DLORENTZ | 590.423.5567       | 2007-02-07 | IT_PROG    |  4200.00 |           0.00 |        103 |            60 |
|         108 | Nancy       | Greenberg   | NGREENBE | 515.124.4569       | 2002-08-17 | FI_MGR     | 12008.00 |           0.00 |        101 |           100 |
|         109 | Daniel      | Faviet      | DFAVIET  | 515.124.4169       | 2002-08-16 | FI_ACCOUNT |  9000.00 |           0.00 |        108 |           100 |
......
|         206 | William     | Gietz       | WGIETZ   | 515.123.8181       | 2002-06-07 | AC_ACCOUNT |  8300.00 |           0.00 |        205 |           110 |
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+

View the table

Sample Solution:

-- Selecting 'department_id'
SELECT department_id 
-- Specifying the table to retrieve data from ('employees')
FROM employees 
-- Filtering the results to include only those with a non-null 'commission_pct'
WHERE commission_pct IS NOT NULL
-- Grouping the results by 'department_id'
GROUP BY department_id 
-- Filtering the grouped results to include only those with a count of 'commission_pct' greater than 10
HAVING COUNT(commission_pct) > 10;

Sample Output:

 department_id
---------------
            80
            50
(2 rows)

N.B.: For necessary region structure of the table have changed. Result may vary.

Code Explanation:

The said query in SQL that selects the department_id of employees who have a commission and groups them by department_id. It then only returns departments where the count of employees with a commission_pct is greater than 10.

Relational Algebra Expression: