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  • 579. 查询员工的累计薪水

    Employee 表保存了一年内的薪水信息。

    请你编写 SQL 语句,来查询一个员工三个月内的累计薪水,但是不包括最近一个月的薪水。

    结果请按 'Id' 升序,然后按 'Month' 降序显示。

    示例:
    输入:

    | Id | Month | Salary |
    |----|-------|--------|
    | 1 | 1 | 20 |
    | 2 | 1 | 20 |
    | 1 | 2 | 30 |
    | 2 | 2 | 30 |
    | 3 | 2 | 40 |
    | 1 | 3 | 40 |
    | 3 | 3 | 60 |
    | 1 | 4 | 60 |
    | 3 | 4 | 70 |
    输出:

    | Id | Month | Salary |
    |----|-------|--------|
    | 1 | 3 | 90 |
    | 1 | 2 | 50 |
    | 1 | 1 | 20 |
    | 2 | 1 | 20 |
    | 3 | 3 | 100 |
    | 3 | 2 | 40 |
     

    解释:

    员工 '1' 除去最近一个月(月份 '4'),有三个月的薪水记录:月份 '3' 薪水为 40,月份 '2' 薪水为 30,月份 '1' 薪水为 20。

    所以近 3 个月的薪水累计分别为 (40 + 30 + 20) = 90,(30 + 20) = 50 和 20。

    | Id | Month | Salary |
    |----|-------|--------|
    | 1 | 3 | 90 |
    | 1 | 2 | 50 |
    | 1 | 1 | 20 |
    员工 '2' 除去最近的一个月(月份 '2')的话,只有月份 '1' 这一个月的薪水记录。

    | Id | Month | Salary |
    |----|-------|--------|
    | 2 | 1 | 20 |
    员工 '3' 除去最近一个月(月份 '4')后有两个月,分别为:月份 '4' 薪水为 60 和 月份 '2' 薪水为 40。所以各月的累计情况如下:

    | Id | Month | Salary |
    |----|-------|--------|
    | 3 | 3 | 100 |
    | 3 | 2 | 40 |

    来源:力扣(LeetCode)
    链接:https://leetcode-cn.com/problems/find-cumulative-salary-of-an-employee
    著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

    leetcode 数据库题目全部题解

    解法一
    首先要明确“三个月内”是什么意思?

    设当前月为a,则a,a-1,a-2为“三个月内”。

    这种“前缀和”问题,通常采用自连接。

    先找出每个月的“三个月内”的全部月份。

    SELECT *
    FROM employee AS E1 JOIN employee AS E2 ON (E1.Id = E2.Id AND E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 3)
    其中:

    E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 3
    表示E1.MONTH的“三个月内”的全部月份。

    再计算每个月三个月内的薪水累加和。

    group by按id和month分组。计算E2.Salary的累加和。结果命名为A。

    (
    SELECT E1.Id,E1.MONTH,SUM(E2.Salary) AS `tsalary`
    FROM employee AS E1 JOIN employee AS E2 ON (E1.Id = E2.Id AND E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 2)
    GROUP BY E1.Id,E1.MONTH
    ) AS A
    又因为不能包括最近一个月的薪水。

    其实是要排除最高月份的值。

    每个员工的最高月份。group by按id分组,最大月份值。

    结果命名为B。

    (
    SELECT E.Id, MAX(E.MONTH) AS `max_month`
    FROM employee AS E
    GROUP BY E.Id
    ) AS B
    结合表A和表B,选出每个员工最高月份之前的值。

    结果请按 ‘Id’ 升序,按 ‘Month’ 降序显示 。

    SELECT A.id,A.month,A.tsalary AS `Salary`
    from
    (
    SELECT E1.Id,E1.MONTH,SUM(E2.Salary) AS `tsalary`
    FROM employee AS E1 JOIN employee AS E2 ON (E1.Id = E2.Id AND E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 2)
    GROUP BY E1.Id,E1.MONTH
    ) AS A
    JOIN
    (
    SELECT E.Id, MAX(E.MONTH) AS `max_month`
    FROM employee AS E
    GROUP BY E.Id
    ) AS B
    ON (A.Id=B.Id AND A.MONTH < B.max_month)
    order by A.id,A.month desc
    解法二
    与解法一思路相同,既然只需要连续三个月的值。

    干脆三个员工表自连接即可。

    又由于1月份,2月份,向前不可能有三个月。

    必然要用left join。

    SELECT *
    FROM employee AS E1
    LEFT JOIN employee AS E2 ON (E1.Id = E2.Id AND E2.MONTH = E1.MONTH -1)
    LEFT JOIN employee AS E3 ON (E3.Id = E2.Id AND E3.MONTH = E2.MONTH -1)
    再讲三个月的薪水相加。注意到E2.Salary和E3.Salary为NULL时,薪水值为0.

    结果命名为A。

    FROM
    (
    SELECT E1.Id,E1.MONTH,(E1.Salary + if(E2.Salary IS NULL,0,E2.Salary) + if(E3.Salary IS NULL,0,E3.Salary)) AS `tsalary`
    FROM employee AS E1
    LEFT JOIN employee AS E2 ON (E1.Id = E2.Id AND E2.MONTH = E1.MONTH -1)
    LEFT JOIN employee AS E3 ON (E3.Id = E2.Id AND E3.MONTH = E2.MONTH -1)
    ) AS A
    求最大月份如解法一。

    (
    SELECT E.Id, MAX(E.MONTH) AS `max_month`
    FROM employee AS E
    GROUP BY E.Id
    ) AS B
    之后同解法一,连接表A和表B,过滤掉最大月份值。

    SELECT A.id,A.month,A.tsalary AS `Salary`
    FROM
    (
    SELECT E1.Id,E1.MONTH,(E1.Salary + if(E2.Salary IS NULL,0,E2.Salary) + if(E3.Salary IS NULL,0,E3.Salary)) AS `tsalary`
    FROM employee AS E1
    LEFT JOIN employee AS E2 ON (E1.Id = E2.Id AND E2.MONTH = E1.MONTH -1)
    LEFT JOIN employee AS E3 ON (E3.Id = E2.Id AND E3.MONTH = E2.MONTH -1)
    ) AS A
    JOIN
    (
    SELECT E.Id, MAX(E.MONTH) AS `max_month`
    FROM employee AS E
    GROUP BY E.Id
    ) AS B
    ON (A.Id=B.Id AND A.MONTH < B.max_month)
    order by A.id,A.month desc
    解法三
    员工表按id升序,再按salary升序。

    应用用户变量,可计算三个月内薪水累加和。

    定义用户变量:@pre_id——上一行的id,@pre_salary1——前一个月的薪水值,@pre_salary2——前两个月的薪水值。

    (SELECT @pre_id:= NULL,@pre_salary1:=0,@pre_salary2:=0) AS T
    每个人,三个月内累计薪水的逻辑:

    if (员工id != @pre_id){
    三个月内累计薪水 = 员工id的薪水
    }else{
    三个月内累计薪水 = 员工id的薪水 + @pre_salary1 + @pre_salary2
    }
    SQL代码:

    IF(E.Id != @pre_id,
    E.Salary
    ,
    E.Salary + @pre_salary1 + @pre_salary2
    ) AS `Salary`
    用户变量更新逻辑:

    if (员工id != @pre_id)
    {
    @pre_salary2 = 0
    }
    else{
    @pre_salary2 = @pre_salary1
    }

    @pre_salary1 = 员工id的薪水
    @pre_id = 员工id的id。
    SQL代码:

    IF(E.Id != @pre_id,
    E.Salary
    ,
    E.Salary + @pre_salary1 + @pre_salary2
    ) AS `Salary`,
    if(E.Id != @pre_id,
    @pre_salary2:=0
    ,
    @pre_salary2:=@pre_salary1
    ) AS `pre2`,
    (@pre_salary1:=E.Salary) AS `pre1`,
    @pre_id:=E.Id
    注意:上述逻辑成立的前提—— 员工表按id升序,再按salary升序。

    将上述逻辑合起来,结果命名为表A。

    (
    SELECT
    E.Id,
    E.MONTH,
    IF(E.Id != @pre_id,
    E.Salary
    ,
    E.Salary + @pre_salary1 + @pre_salary2
    ) AS `Salary`,
    if(E.Id != @pre_id,
    @pre_salary2:=0
    ,
    @pre_salary2:=@pre_salary1
    ) AS `pre2`,
    (@pre_salary1:=E.Salary) AS `pre1`,
    @pre_id:=E.Id
    FROM employee AS E,
    (SELECT @pre_id:= NULL,@pre_salary1:=0,@pre_salary2:=0) AS T
    ORDER BY E.Id,E.MONTH
    ) AS A
    每个员工的最大月份类似解法一。

    (
    SELECT E.Id, MAX(E.MONTH) AS `max_month`
    FROM employee AS E
    GROUP BY E.Id
    ) AS B
    之后同解法一,连接表A和表B,过滤掉最大月份值。

    SELECT A.Id,A.MONTH,A.salary
    FROM
    (
    SELECT
    E.Id,
    E.MONTH,
    IF(E.Id != @pre_id,
    E.Salary
    ,
    E.Salary + @pre_salary1 + @pre_salary2
    ) AS `Salary`,
    if(E.Id != @pre_id,
    @pre_salary2:=0
    ,
    @pre_salary2:=@pre_salary1
    ) AS `pre2`,
    (@pre_salary1:=E.Salary) AS `pre1`,
    @pre_id:=E.Id
    FROM employee AS E,
    (SELECT @pre_id:= NULL,@pre_salary1:=0,@pre_salary2:=0) AS T
    ORDER BY E.Id,E.MONTH
    ) AS A
    JOIN
    (
    SELECT E.Id, MAX(E.MONTH) AS `max_month`
    FROM employee AS E
    GROUP BY E.Id
    ) AS B
    ON (A.Id=B.Id AND A.MONTH < B.max_month)
    order by A.id,A.month desc

    作者:jason-2
    链接:https://leetcode-cn.com/problems/find-cumulative-salary-of-an-employee/solution/san-chong-jie-fa-by-jason-2/
    来源:力扣(LeetCode)
    著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

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  • 原文地址:https://www.cnblogs.com/leeeee/p/11902035.html
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