mysql 取每个部门薪资最高的1人

员工表：
QQ图片20180716142827.png

部门表：
QQ图片20180716144304.png

要求：取每个部门工资最高的1人，如图：
捕获.PNG

创建Employee(员工表)，Department(部门表)

SQL 1:

SELECT
e.id,
d. NAME dName,
e. NAME eName,
e.salary
FROM
(
SELECT
*
FROM
Employee
ORDER BY
salary DESC
) e
LEFT JOIN Department d ON e.departmentid = d.id
GROUP BY
e.departmentid

SQL 2:

SELECT
e.id,
d. NAME dName,
e. NAME eName,
e.salary
FROM
Employee e
LEFT JOIN Department d ON e.departmentid = d.id
WHERE
salary = (
SELECT
max(salary)
FROM
Employee
WHERE
departmentid = e.departmentid
)
ORDER BY
salary DESC

SQL 3:
…
SQL 4:
…

方法很多…