-3

我有一张这样的桌子:

雇员

EmpID | EmpName |EmpSalary | DEpartment

然后是另一张桌子

奖金

EmpID |EmpBonus

如何获得每个部门最高薪的员工,包括他们的奖金?

4

3 回答 3

4

如果您使用的是 SQL-Server,请使用ROW_NUMBERDENSE_RANK(如果您想包含关联):

WITH CTE AS(
    SELECT e.EmpID,e.EmpName,e.EmpSalary, e.Department,b.EmpBonus,
       RN = ROW_NUMBER() OVER (PARTITION BY Department 
                               ORDER BY (EmpSalary + COALESCE(EmpBonus,0)) DESC)
    FROM Employees e LEFT OUTER JOIN Bonuses b
       ON e.EmpID = b.EmpID 
)
SELECT EmpID, EmpName, EmpSalary, Department, EmpBonus
FROM CTE
WHERE RN = 1

排名功能

于 2013-09-17T09:07:30.783 回答
-1

在 SQL Server 中

SELECT top 1 e.EmpID,e.EmpName,e.Department,
sum(EmpSalary + EmpBonus) as total_salary
FROM Employees e INNER JOIN Bonuses b
ON e.EmpID = b.EmpID 
group by e.EmpID,e.EmpName,e.Department
order by total_salary desc

在 MySQL 中

SELECT e.EmpID,e.EmpName,e.Department,
sum(EmpSalary + EmpBonus) as total_salary
FROM Employees e INNER JOIN Bonuses b
ON e.EmpID = b.EmpID 
group by e.EmpID,e.EmpName,e.Department
order by total_salary desc limit 1
于 2013-09-17T09:12:06.907 回答
-1

尝试这个:

select EmpID (EmpSalary+EmpBonus) as total_salary
from Employees e LEFT JOIN Bonuses b
order by total_salary desc
limit 1

我不知道您使用的是哪个 SGBD,因此限制 1 可能是错误的。按照你的情况去做。

于 2013-09-17T09:13:28.063 回答