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MySQL Aggregate Function Exercises: List the number of jobs available in the employees table

MySQL Aggregate Function: Exercise-1 with Solution

Write a query to list the number of jobs available in the employees table.

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 distinct job IDs in the employees table
SELECT COUNT(DISTINCT job_id) 
-- Selecting data from the employees table
FROM employees;

Explanation:

  • This SQL query counts the number of distinct job IDs present in the job_id column of the employees table.
  • The COUNT(DISTINCT job_id) function calculates the number of unique job IDs.
  • The DISTINCT keyword ensures that each job ID is only counted once, even if it appears multiple times in the table.
  • This query is useful when you want to find out how many different types of jobs are represented among the employees in the organization.

Relational Algebra Expression:

Relational Algebra Expression: Aggregate Function: List the number of jobs available in the employees table.


Relational Algebra Tree:

Relational Algebra Tree: Aggregate Function: List the number of jobs available in the employees table.


Pictorial Presentation of the above query

Pictorial: Query to list the number of jobs available in the employees table.


Go to:


PREV :Aggregate Functions and Group by
NEXT : Write a query to get the total salaries payable to employees.

MySQL Code Editor:

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