用于DISTINCT ON更简单、更快速的查询，可以满足您的所有要求：
SELECT DISTINCT ON (d.id)
d.id AS department_id, d.name AS department
,e.id AS employee_id, e.name AS employee, e.salary
FROM departments d
LEFT JOIN employees e ON e.department_id = d.id
ORDER BY d.id, e.salary DESC;
->SQLfiddle（用于 Postgres）。
另请注意LEFT [OUTER] JOIN，结果中没有员工的部门。
这只会选择one每个部门的员工。如果有多个共享最高薪水，您可以添加更多 ORDER BY 项目以特别选择一个。否则，从同行中选择任意一个。
如果没有员工，仍会列出部门，其中NULL包含员工列的值。
您可以简单地在列表中添加您需要的任何列SELECT。
在此相关答案中找到该技术的详细解释、链接和基准：
选择每个 GROUP BY 组中的第一行？
旁白：使用非描述性的列名（如nameor ）是一种反模式id。应该是employee_id等employee
使用窗口函数rank()（就像@Scotch 已经发布的一样，更简单、更快）：
SELECT d.name AS department, e.employee, e.salary
FROM departments d
LEFT JOIN (
SELECT name AS employee, salary, department_id
,rank() OVER (PARTITION BY department_id ORDER BY salary DESC) AS rnk
FROM employees e
) e ON e.department_id = d.department_id AND e.rnk = 1;
与您的示例（没有联系）的上述查询相同的结果，只是慢了一点。
这基本上就是你想要的。Rank() Over
SELECT ename ,
departments.name
FROM ( SELECT ename ,
dname
FROM ( SELECT employees.name as ename ,
departments.name as dname ,
rank() over (
PARTITION BY employees.department_id
ORDER BY employees.salary DESC
)
FROM Employees
JOIN Departments on employees.department_id = departments.id
) t
WHERE rank = 1
) s
RIGHT JOIN departments on s.dname = departments.name
这是参考你的小提琴：
SELECT * -- or whatever is your columns list.
FROM employees e JOIN departments d ON e.Department_ID = d.id
WHERE (e.Department_ID, e.Salary) IN (SELECT Department_ID, MAX(Salary)
FROM employees
GROUP BY Department_ID)
编辑 ：
正如下面评论中提到的，如果您还想查看 IT 部门，所有NULL员工记录，您可以使用RIGHT JOINand 将过滤条件放在加入子句本身中，如下所示：
SELECT e.name, e.salary, d.name -- or whatever is your columns list.
FROM employees e RIGHT JOIN departments d ON e.Department_ID = d.id
AND (e.Department_ID, e.Salary) IN (SELECT Department_ID, MAX(Salary)
FROM employees
GROUP BY Department_ID)
SELECT
e.first_name, d.department_name, e.salary
FROM
employees e
JOIN
departments d
ON
(e.department_id = d.department_id)
WHERE
e.first_name
IN
(SELECT TOP 2
first_name
FROM
employees
WHERE
department_id = d.department_id);
好老的经典sql：
select e1.name, e1.salary, e1.department_id
from employees e1
where e1.salary=
(select maxsalary=max(e.salary) --, e. department_id
from employees e
where e.department_id = e1.department_id
group by e.department_id
)
`select d.Name, e.Name, e.Salary from Employees e, Departments d,
(select DepartmentId as DeptId, max(Salary) as Salary
from Employees e
group by DepartmentId) m
where m.Salary = e.Salary
and m.DeptId = e.DepartmentId
and e.DepartmentId = d.DepartmentId`
每个部门的最高工资在内部查询中使用 GROUP BY 计算。然后选择满足这些限制的员工。
表 1 是 emp - empno、ename、sal、deptno
表 2 是 dept - deptno, dname。
select e1.empno, e1.ename, e1.sal, e1.deptno as department
from emp e1
where e1.sal in
(SELECT max(sal) from emp e, dept d where e.deptno = d.deptno group by d.dname)
order by e1.deptno asc;
SQL查询：
select d.name,e.name,e.salary
from employees e, depts d
where e.dept_id = d.id
and (d.id,e.salary) in
(select dept_id,max(salary) from employees group by dept_id);
假设 Postgres
返回带有员工详细信息的最高薪水，假设表名 emp 的员工部门为 dept_id
select e1.* from emp e1 inner join (select max(sal) avg_sal,dept_id from emp group by dept_id) as e2 on e1.dept_id=e2.dept_id and e1.sal=e2.avg_sal
返回每个部门工资最高的一个或多个人：
SELECT result.Name Department, Employee2.Name Employee, result.salary Salary
FROM ( SELECT dept.name, dept.department_id, max(Employee1.salary) salary
FROM Departments dept
JOIN Employees Employee1 ON Employee1.department_id = dept.department_id
GROUP BY dept.name, dept.department_id ) result
JOIN Employees Employee2 ON Employee2.department_id = result.department_id
WHERE Employee2.salary = result.salary
看看这个解决方案 SELECT MAX(E.SALARY), E.NAME, D.NAME as Department FROM employees E INNER JOIN DEPARTMENTS D ON D.ID = E.DEPARTMENT_ID GROUP BY D.NAME