Question

我今天在Hacker News上发现了几个SQL tasks，但是我坚持要解决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：

enter image description here

列出各自部门薪水最高的员工。

我为你设置了一个SQL小提琴here。应该归还特里罗宾逊，劳拉怀特。除了他们的名字，它应该有他们的薪水和部门名称。

此外，我很想知道将返回Terry Robinsons（销售部门的最高薪水）和Laura White（营销部门的最高薪水）的查询以及IT部门的空行，以null为员工;明确说明该部门没有雇员（因此没有薪水最高的人）。

Answer 1

返回一名员工，每个部门的工资最高。

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（对于Postgres）。

另请注意LEFT [OUTER] JOIN使部门在结果中没有员工。

每个部门只挑选one名员工。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，从同行中挑选出任意一个如果没有员工，则仍会列出部门，其中员工列的值为NULL。

您只需在SELECT列表中添加所需的任何列。

在此相关答案中查找该技术的详细说明，链接和基准：
Select first row in each GROUP BY group?

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

返回所有员工，每个部门的工资最高。

使用窗口函数rank()（如@Scotch already posted，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（没有关系）的上述查询结果相同，只是稍微慢一些。

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

这是关于你的小提琴：

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：

如下面的评论所述，如果您想要查看IT部门，并且员工记录的所有NULL都可以使用RIGHT JOIN并将过滤条件放入加入条款中本身如下：

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

为薪水最高的每个部门返回一个或多个人：

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 5

好古老的经典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 6

表1是emp - empno，ename，sal，deptno

表2是dept - deptno，dname。

查询可以是（包括11.2g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 7

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 8

`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`

使用GROUP BY在内部查询中计算每个部门的最高工资。然后选择满足这些约束条件的员工。

Answer 9

假设Postgres

返回员工详细信息的最高薪水，假设表名为emp且员工部门为dept_id

df1 = df[((df['category'] == 'A') & (df['value'].between(10,20))) | 
         (df['category'] != 'A')]
print (df1)
  category  value
1        B     10
2        A     15
3        B     28
4        A     18

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 MAX（E.SALARY） E.NAME， D.NAME作为部门来自员工E. 内部联合部门D ON D.ID = E.DEPARTMENT_ID GROUP BY D.NAME

部门薪水最高的员工

11 个答案:

返回一名员工，每个部门的工资最高。

返回所有员工，每个部门的工资最高。

查询可以是（包括11.2g的关系和运行）：