Question

我在mysql中有表A

+-------------------------+
| ID  Name Month  Salary  |
+-------------------------+
| 1  Joe   4      300     |
| 2  Kim   3     100      |
| 3  Moko  2     150      |
| 4   Loli  5     80      |
| 5  Tom   3    400       |
+-------------------------+

我如何获得基于月薪和最高收入者的工资总额

示例

Joe = 4*300 = 1200
Kim = 300
Moko = 300
Loli = 400
Tom = 1200

输出将：1200，2

最高薪水为1200，而有薪水的人为Joe和Tom，为

Answer 1

尝试使用子查询，如下所示

    select max(salary),count(*) from (select name, sum(Month*Salary) as salary
    from data    group by name
    ) b where b.salary= 

   (select max(salary) from
   (select sum(Month*Salary) as salary
    from data    group by name
    )b
    )

Demo link

Answer 2

您可以计算子查询中的所有total工资，然后计算每个total值的出现次数，按total降序为LIMIT 1到限制为最高值：

SELECT total, COUNT(*) AS num
FROM (
  SELECT Month * Salary AS total
  FROM data
) t
GROUP BY total
ORDER BY total DESC
LIMIT 1

输出：

total   num
1200    2

Demo on SQLFiddle

Answer 3

SELECT Month*Salary, COUNT(*)
FROM A WHERE Month*Salary = (SELECT MAX(Month*Salary)
                             FROM A)

第二个查询将为您提供最大的薪水总额，然后您可以筛选表A中的月*薪水等于该金额的条目。

sql查询最大累积

3 个答案: