MySQL Aggregate Function Exercises: Get the average salary and number of employees working the department 90

MySQL Aggregate Function: Exercise-5 with Solution

Write a query to get the average salary and number of employees working the department 90.

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:

-- Calculating the average salary and counting the number of employees in department 90
SELECT AVG(salary), COUNT(*) 
-- Selecting data from the employees table
FROM employees 
-- Filtering the result set to include only employees in department 90
WHERE department_id = 90;

Explanation:

• This SQL query calculates the average salary and counts the number of employees in department 90.
• The AVG() function calculates the average value of the specified column (salary in this case).
• The COUNT(*) function counts the number of rows in the table, which effectively counts the total number of employees.
• The WHERE clause filters the result set to include only employees who belong to department 90.
• This query is useful when you want to find out the average salary and the number of employees in a specific department, in this case, department 90.

Relational Algebra Expression:

Relational Algebra Tree:

Pictorial Presentation of the above query

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