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Table: Employees
+-------------+---------+ | Column Name | Type | +-------------+---------+ | employee_id | int | | name | varchar | | salary | int | +-------------+---------+ employee_id is the primary key (column with unique values) for this table. Each row of this table indicates the employee ID, employee name, and salary.
Write a solution to calculate the bonus of each employee. The bonus of an employee is 100% of their salary if the ID of the employee is an odd number and the employee's name does not start with the character 'M'. The bonus of an employee is 0 otherwise.
Return the result table ordered by employee_id.
The result format is in the following example.
Example 1:
Input: Employees table: +-------------+---------+--------+ | employee_id | name | salary | +-------------+---------+--------+ | 2 | Meir | 3000 | | 3 | Michael | 3800 | | 7 | Addilyn | 7400 | | 8 | Juan | 6100 | | 9 | Kannon | 7700 | +-------------+---------+--------+ Output: +-------------+-------+ | employee_id | bonus | +-------------+-------+ | 2 | 0 | | 3 | 0 | | 7 | 7400 | | 8 | 0 | | 9 | 7700 | +-------------+-------+ Explanation: The employees with IDs 2 and 8 get 0 bonus because they have an even employee_id. The employee with ID 3 gets 0 bonus because their name starts with 'M'. The rest of the employees get a 100% bonus.
We can use the IF statement to determine the calculation method of the bonus, and then use ORDER BY to sort the results by employee_id.
# Write your MySQL query statement below
SELECT
employee_id,
IF(
employee_id % 2 = 0
OR LEFT(name, 1) = 'M',
0,
salary
) AS bonus
FROM
employees
ORDER BY 1;