SQL Exercises: Employees who was hired after the employee with ID 165

Last update on March 12 2025 12:11:50 (UTC/GMT +8 hours)

From the following table, write a SQL query to find those employees who joined after the employee whose ID is 165. Return first name, last name and hire date.

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 a concatenated string of 'first_name' and 'last_name' as 'Full_Name', and the 'hire_date' from the 'employees' table
SELECT first_name ||' '|| last_name AS Full_Name , hire_date 
	
-- Filtering rows based on the condition that the 'hire_date' is greater than the 'hire_date' of the employee with 'employee_id' 165 in the 'employees' table
FROM employees 

-- Subquery to find the 'hire_date' where the 'employee_id' is 165 in the 'employees' table
WHERE hire_date > (
                     SELECT hire_date 
                       FROM employees 
                        WHERE employee_id = 165);

Sample Output:

full_name	hire_date
Steven Markle	2008-03-08
Sundar Ande	2008-03-24
Amit Banda	2008-04-21
Sundita Kumar	2008-04-21

Code Explanation:

The said query in SQL that retrieves the first name and last name concatenated as "Full_Name" and the hire date from the 'employees' table, where the hire date is greater than the hire date of the employee with an "employee_id" of 165.
The "WHERE" clause is used to filter the results so that only those records are retrieved where the "hire_date" is greater than the "hire_date" of the employee with "employee_id" 165, which is retrieved using a sub-query.

Visual Presentation:

Alternative Solutions:

Using a JOIN with > Operator:

SELECT e1.first_name || ' ' || e1.last_name AS Full_Name, e1.hire_date 
FROM employees e1
JOIN employees e2 ON e1.hire_date > e2.hire_date AND e2.employee_id = 165;

Using Subquery with a Comparison Operator:

SELECT first_name || ' ' || last_name AS Full_Name, hire_date 
FROM employees 
WHERE hire_date > (
    SELECT MAX(hire_date) 
    FROM employees 
    WHERE employee_id = 165
);

Using EXISTS and a Subquery:

SELECT first_name || ' ' || last_name AS Full_Name, hire_date 
FROM employees e1
WHERE EXISTS (
    SELECT 1
    FROM employees e2
    WHERE e2.employee_id = 165 AND e1.hire_date > e2.hire_date
);

Practice Online

Query Visualization:

Duration:

Rows:

Cost:

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

Previous SQL Exercise: Employees earning more than departments minimum wage.
Next SQL Exercise: Minimum salary of a department which ID is 70.