SQL Exercises: Display the details of departments managed by Susan

Last update on March 12 2025 12:10:49 (UTC/GMT +8 hours)

From the following table, write a SQL query to find the departments managed by Susan. Return all the fields of departments.

Sample table: departments

+---------------+----------------------+------------+-------------+
| DEPARTMENT_ID | DEPARTMENT_NAME      | MANAGER_ID | LOCATION_ID |
+---------------+----------------------+------------+-------------+
|            10 | Administration       |        200 |        1700 |
|            20 | Marketing            |        201 |        1800 |
|            30 | Purchasing           |        114 |        1700 |
|            40 | Human Resources      |        203 |        2400 |
|            50 | Shipping             |        121 |        1500 |
|            60 | IT                   |        103 |        1400 |
|            70 | Public Relations     |        204 |        2700 |
|            80 | Sales                |        145 |        2500 |
|            90 | Executive            |        100 |        1700 |
|           100 | Finance              |        108 |        1700 |
......
+---------------+----------------------+------------+-------------+

View the 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       | 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 all columns (*) from the 'departments' table
SELECT * 

-- Filtering rows from the 'departments' table based on the condition that the 'manager_id' is in the result set of a subquery
FROM departments 

-- Subquery to find 'employee_id' values where the 'first_name' is 'Susan' in the 'employees' table
WHERE manager_id IN 
	(SELECT employee_id 
	 FROM employees 
	 WHERE first_name='Susan');

Sample Output:

department_id	department_name	manager_id	location_id
40		Human Resources	203 		2400

Code Explanation:

The said query in SQL that selects all the rows from the 'departments' table where the "manager_id" column matches any of the values returned by the subquery. The subquery selects the "employee_id" values from the 'employees' table where the "first_name" column equals 'Susan'.

Visual Presentation:

Alternative Solutions:

Using JOINs:

SELECT d.*
FROM departments d
JOIN employees e ON d.manager_id = e.employee_id
WHERE e.first_name = 'Susan';

Using EXISTS:

SELECT d.*
FROM departments d
WHERE EXISTS (
    SELECT 1
    FROM employees e
    WHERE e.employee_id = d.manager_id
    AND e.first_name = 'Susan'
);

Using Scalar Subquery in the WHERE Clause:

SELECT * 
FROM departments 
WHERE manager_id = (
    SELECT employee_id 
    FROM employees 
    WHERE first_name = 'Susan'
);

Practice Online

Query Visualization:

Duration:

Rows:

Cost:

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