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  • [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
    

      

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