部门工资最高的员工

Question

我今天在 Hacker News 上发现了几个 SQL tasks，但是我一直在解决 Postgres 中的第二个任务，我将在这里描述：

您有以下简单的表结构：

列出各自部门薪水最高的员工。

我设置了一个 SQL Fiddle here 供您使用。它应该返回特里罗宾逊，劳拉怀特。除了他们的名字，还应该有他们的薪水和部门名称。

此外，我很想知道一个查询会返回 Terry Robinsons（销售部门的最高薪水）和 Laura White（营销部门的最高薪水）以及 IT 部门的空行，并带有 @ 987654324@作为员工；明确说明该部门没有员工（因此没有人工资最高）。

Answer 1

返回一个每个部门工资最高的员工。

使用DISTINCT ON 进行更简单、更快速的查询，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，它会在结果中保留没有员工的部门。

这仅选择每个部门的 one 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，会从同行中任意挑选一个。
如果没有员工，仍会列出部门，其中员工列的值为 NULL。

您可以简单地在SELECT 列表中添加您需要的任何列。

在此相关答案中找到该技术的详细说明、链接和基准：
Select first row in each GROUP BY group?

除此之外：使用非描述性列名（如 name 或 id）是一种反模式。应该是employee_id、employee等。

返回所有个每个部门薪水最高的员工。

使用窗口函数rank()（类似@Scotch already posted，只是更简单更快）：

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（没有关联）的上述查询结果相同，只是慢了一点。

Answer 2

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

Answer 3

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 用于员工记录，您可以使用RIGHT JOIN 并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)

Answer 4

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
   )

Answer 5

Table1 是 emp - empno、ename、sal、deptno

Table2 是 dept - deptno, dname。

查询可能是（包括 11.2g 上的关系和运行）：

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; 

Answer 6

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);

Answer 7

`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是在内部查询中使用 GROUP BY 计算的。然后选择满足这些限制条件的员工。

Answer 8

假设 Postgres

返回带有员工详细信息的最高薪水，假设表名 emp 具有带 dept_id 的员工部门

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

Answer 9

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

Answer 10

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);

Answer 11

看看这个解决方案 选择 最大（E.SALARY）， 姓名， D.NAME 作为部门 来自员工 E 内部连接 ​​D.ID = E.DEPARTMENT_ID 上的 DEPARTMENTS 按 D.NAME 分组