SQL Exercise: Display all the details of managers

SQL subqueries on employee Database: Exercise-1 with Solution

[An editor is available at the bottom of the page to write and execute the scripts.]

1. From the following table, write a SQL query to find the managers. Return complete information about the managers.

Sample table: employees

Sample Solution:

SELECT *
FROM employees
WHERE emp_id IN
    (SELECT manager_id
     FROM employees);

Sample Output:

 emp_id | emp_name | job_name  | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+-----------+------------+------------+---------+------------+--------
  68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001
  66928 | BLAZE    | MANAGER   |      68319 | 1991-05-01 | 2750.00 |            |   3001
  67832 | CLARE    | MANAGER   |      68319 | 1991-06-09 | 2550.00 |            |   1001
  65646 | JONAS    | MANAGER   |      68319 | 1991-04-02 | 2957.00 |            |   2001
  67858 | SCARLET  | ANALYST   |      65646 | 1997-04-19 | 3100.00 |            |   2001
  69062 | FRANK    | ANALYST   |      65646 | 1991-12-03 | 3100.00 |            |   2001
(6 rows)

Explanation:

The said query in SQL that selects all columns from the 'employees' table where the "emp_id" value is found in the results of a subquery. The subquery selects the "manager_id" values from the 'employees' table. The query likely returns a list of employees who are also managers.

Alternative Solutions:

Using EXISTS:

-- This query selects all columns from the 'employees' table, aliasing it as 'e'
SELECT *
-- It filters the results to include only those rows where there exists a record in the subquery
FROM employees e
-- The subquery checks for the existence of at least one record where 'manager_id' matches 'emp_id' from the outer query
WHERE EXISTS (
    SELECT 1
    FROM employees
    WHERE manager_id = e.emp_id
);
-- The main query selects all records from 'employees' where there exists a matching 'manager_id' for the current 'emp_id'

Explanation:

This query uses the EXISTS clause to check for the existence of rows in the subquery where the manager_id matches the emp_id of the outer query.

Using a Correlated Subquery:

-- This query selects all columns from the 'employees' table, aliasing it as 'e'
SELECT *
-- It filters the results to include only those rows where 'emp_id' is equal to ANY value in the subquery
FROM employees e
-- The subquery selects 'manager_id' values from the 'employees' table, excluding NULL values
WHERE e.emp_id = ANY (
    SELECT manager_id
    FROM employees
    WHERE manager_id IS NOT NULL
);
-- The main query selects all records from 'employees' where 'emp_id' matches ANY 'manager_id' value from the subquery

Explanation:

This query uses a correlated subquery with the ANY operator to compare each emp_id in the outer query with the manager_id in the subquery, returning rows where a match is found.

Practice Online

Structure of employee Database

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous SQL Exercise: SQL Subqueries - Exercises, Practice, Solution
Next SQL Exercise: Display details and experience of all the managers.