SQL Exercises: Department where the employee works with ID 201

Last update on March 12 2025 12:11:53 (UTC/GMT +8 hours)

From the following table, write a SQL query to find employees who work for the department in which employee ID 201 is employed. Return first name, last name, salary, and 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 |
|         110 | John        | Chen        | JCHEN    | 515.124.4269       | 2005-09-28 | FI_ACCOUNT |  8200.00 |           0.00 |        108 |           100 |
....................
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+

View the table

Sample Solution:

-- Selecting specific columns (first_name, last_name, salary, department_id) from the 'employees' table
SELECT first_name, last_name, salary, department_id 

-- Filtering rows based on the condition that the 'department_id' matches the result of a subquery
FROM employees 

-- Subquery to find the 'department_id' where the 'employee_id' is 201 in the 'employees' table
WHERE department_id = 
     (SELECT department_id 
       FROM employees 
        WHERE employee_id = 201);

Sample Output:

first_name	last_name	salary	department_id
Michael		Hartstein	13000.00	20
Pat		Fay		6000.00		20

Code Explanation:

The said query in SQL that retrieves the first name, last name, salary, and department id of all employees who belong to the same department as employee id 201.
The WHERE clause specifies that checking if the department_id of each employee matches the department_id of employee id 201. The condition uses a subquery in the WHERE clause to retrieve the department_id of employee id 201, and then compare it to the department_id of each employee.

Visual Presentation:

Alternative Solutions:

Using JOIN:

SELECT e1.first_name, e1.last_name, e1.salary, e1.department_id 
FROM employees e1
JOIN employees e2 ON e1.department_id = e2.department_id
WHERE e2.employee_id = 201;

Using EXISTS:

SELECT first_name, last_name, salary, department_id 
FROM employees e1
WHERE EXISTS (
    SELECT 1
    FROM employees e2
    WHERE e2.employee_id = 201 AND e1.department_id = e2.department_id
);

Practice Online

Query Visualization:

Duration:

Rows:

Cost:

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