Description
Table: Employees
+-------------+----------+ | Column Name | Type | +-------------+----------+ | employee_id | int | | name | varchar | | manager_id | int | | salary | int | +-------------+----------+ In SQL, employee_id is the primary key for this table. This table contains information about the employees, their salary, and the ID of their manager. Some employees do not have a manager (manager_id is null).
Find the IDs of the employees whose salary is strictly less than $30000 and whose manager left the company. When a manager leaves the company, their information is deleted from the Employees table, but the reports still have their manager_id set to the manager that left.
Return the result table ordered by employee_id.
The result format is in the following example.
Example 1:
Input: Employees table: +-------------+-----------+------------+--------+ | employee_id | name | manager_id | salary | +-------------+-----------+------------+--------+ | 3 | Mila | 9 | 60301 | | 12 | Antonella | null | 31000 | | 13 | Emery | null | 67084 | | 1 | Kalel | 11 | 21241 | | 9 | Mikaela | null | 50937 | | 11 | Joziah | 6 | 28485 | +-------------+-----------+------------+--------+ Output: +-------------+ | employee_id | +-------------+ | 11 | +-------------+ Explanation: The employees with a salary less than $30000 are 1 (Kalel) and 11 (Joziah). Kalel's manager is employee 11, who is still in the company (Joziah). Joziah's manager is employee 6, who left the company because there is no row for employee 6 as it was deleted.
Solution
MySQL
# Write your MySQL query statement below
select e.employee_id
from employees e
where e.salary < 30000 and e.manager_id is not null and not exists (
select 1
from employees e2
where e2.employee_id = e.manager_id
)
order by e.employee_id;