0
0
Read Time:1 Minute, 4 Second
Table: Employees
+-------------+---------+ | Column Name | Type | +-------------+---------+ | employee_id | int | | name | varchar | | salary | int | +-------------+---------+ employee_id is the primary key for this table. Each row of this table indicates the employee ID, employee name, and salary.
Write an SQL query 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 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 query 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.
select employee_id , if(name not like 'M%' AND employee_id%2!=0,salary,0) as bonus
from Employees order by employee_id ;
Happy
0 %
Sad
0 %
Excited
0 %
Sleepy
0 %
Angry
0 %
Surprise
0 %
About Author
