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577. Employee Bonus
Description
Table: Employee
+-------------+---------+ | Column Name | Type | +-------------+---------+ | empId | int | | name | varchar | | supervisor | int | | salary | int | +-------------+---------+ empId is the column with unique values 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 column of unique values for this table. empId is a foreign key (reference column) to empId from the Employee table. Each row of this table contains the id of an employee and their respective bonus.
Write a solution to report the name and bonus amount of each employee with a bonus less than 1000.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Employee table: +-------+--------+------------+--------+ | empId | name | supervisor | salary | +-------+--------+------------+--------+ | 3 | Brad | null | 4000 | | 1 | John | 3 | 1000 | | 2 | Dan | 3 | 2000 | | 4 | Thomas | 3 | 4000 | +-------+--------+------------+--------+ Bonus table: +-------+-------+ | empId | bonus | +-------+-------+ | 2 | 500 | | 4 | 2000 | +-------+-------+ Output: +------+-------+ | name | bonus | +------+-------+ | Brad | null | | John | null | | Dan | 500 | +------+-------+
Solutions
Solution 1: Left Join
We can use a left join to join the Employee table and the Bonus table on empId, and then filter out the employees whose bonus is less than $1000$. Note that the employees with NULL bonus values after the join should also be filtered out, so we need to use the IFNULL function to convert NULL values to $0$.
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# Write your MySQL query statement below SELECT name, bonus FROM Employee LEFT JOIN Bonus USING (empId) WHERE IFNULL(bonus, 0) < 1000;