SQL Exercise: Display details and experience of all the managers

Last update on January 20 2026 12:30:42 (UTC/GMT +8 hours)

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2. From the following table, write a SQL query to compute the experience of all the managers. Return employee ID, employee name, job name, joining date, and experience.

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 |
.........
|         206 | William     | Gietz       | WGIETZ   | 515.123.8181       | 2002-06-07 | AC_ACCOUNT |  8300.00 |           0.00 |        205 |           110 |
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+

View the table

Sample Solution:

SELECT emp_id,
       emp_name,
       job_name,
       hire_date,
       age(CURRENT_DATE, hire_date) "Experience"
FROM employees
WHERE emp_id IN
    (SELECT manager_id
     FROM employees);

Sample Output:

 emp_id | emp_name | job_name  | hire_date  |       Experience
--------+----------+-----------+------------+-------------------------
  68319 | KAYLING  | PRESIDENT | 1991-11-18 | 26 years 2 mons 17 days
  66928 | BLAZE    | MANAGER   | 1991-05-01 | 26 years 9 mons 4 days
  67832 | CLARE    | MANAGER   | 1991-06-09 | 26 years 7 mons 26 days
  65646 | JONAS    | MANAGER   | 1991-04-02 | 26 years 10 mons 3 days
  67858 | SCARLET  | ANALYST   | 1997-04-19 | 20 years 9 mons 16 days
  69062 | FRANK    | ANALYST   | 1991-12-03 | 26 years 2 mons 2 days
(6 rows)

Explanation:

The given query in SQL that returns a table of information about employees who are also managers, including their ID, name, job name, hire date, and experience from the 'employees' table.

The CURRENT_DATE function calculates the "Experience" as the difference between the current date and the "hire_date" of each employee.

The WHERE clause only includes those employees in the result set who are managers.

Alternative Solution:

Using EXISTS:


-- Selecting employee details for those who have a manager
SELECT emp_id,
       emp_name,
       job_name,
       hire_date,
       age(CURRENT_DATE, hire_date) AS "Experience"
FROM employees e
WHERE EXISTS
    (
        -- Subquery checks if there is at least one record
        -- in the "employees" table where the manager_id
        -- matches the emp_id of the outer query's employee
        SELECT 1
        FROM employees
        WHERE manager_id = e.emp_id
    );

Explanation:

The EXISTS subquery is used to check if there exists a record in the employees table where the manager_id matches the emp_id of the outer query's employee, filtering the results accordingly.

Go to:

PREV : Display all the details of managers.
NEXT : Display the list in ascending order of location.

Practice Online

Sample Database: employees

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