[LeetCode]577. Employee Bonus 员工奖金

Select all employee‘s name and bonus whose bonus is < 1000.

Table:Employee

+-------+--------+-----------+--------+
| empId |  name  | supervisor| salary |
+-------+--------+-----------+--------+
|   1   | John   |  3        | 1000   |
|   2   | Dan    |  3        | 2000   |
|   3   | Brad   |  null     | 4000   |
|   4   | Thomas |  3        | 4000   |
+-------+--------+-----------+--------+
empId is the primary key column for this table.

Table: Bonus

+-------+-------+
| empId | bonus |
+-------+-------+
| 2     | 500   |
| 4     | 2000  |
+-------+-------+
empId is the primary key column for this table.

Example ouput:

+-------+-------+
| name  | bonus |
+-------+-------+
| John  | null  |
| Dan   | 500   |
| Brad  | null  |
+-------+-------+

选出所有奖金<1000元的雇员姓名及奖金数额

解法1：

# Write your MySQL query statement below
SELECT name, bonus
FROM Employee LEFT JOIN Bonus USING (empId)
WHERE IFNULL(bonus, 0) < 1000　　

解法2：

select name, bonus
from
Employee e left join Bonus b
on e.empId = b.empId
where bonus < 1000 or bonus is null

原文地址：https://www.cnblogs.com/lightwindy/p/9698931.html