Employee Bonus

Problem

Table: Employee

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| empId       | int     |
| name        | varchar |
| supervisor  | int     |
| salary      | int     |
+-------------+---------+
empId is the primary key column for this table.
Each row of this table indicates the name and the ID of an employee in addition to their salary and the id of their manager.

Table: Bonus

+-------------+------+
| Column Name | Type |
+-------------+------+
| empId       | int  |
| bonus       | int  |
+-------------+------+
empId is the primary key column for this table.
empId is a foreign key to empId from the Employee table.
Each row of this table contains the id of an employee and their respective bonus.

Write an SQL query to report the name and bonus amount of each employee with a bonus less than 1000.

Return the result table in any order.

Solution

The problem Employee Bonus can be solved using the keyword LEFT OUTER JOIN to check if each employee has a bonus less than 1000.

Implementation

# Write your MySQL query statement below

SELECT employee.name AS name, bonus.bonus AS bonus
FROM
  employee
  LEFT OUTER JOIN bonus ON employee.empId = bonus.empId
WHERE bonus.bonus IS NULL OR bonus.bonus < 1000;