mysql - 选择所有部门中薪水最高的所有员工

Question

在其中一次采访中，有人问我以下问题

“写一个查询，找出所有部门的所有员工，该部门的工资最高的部门名称，员工姓名和他的工资”

也就是说，employee 表有 100 条记录，department 表有 10 条记录。所以它需要从查询中给我 10 条记录，加上如果任何部门都没有员工，它仍然需要显示该部门名称。

谢谢

Answer 1

在没有看到表结构的情况下，我会说您可能可以通过几种不同的方式来做到这一点。

使用IN子句：

select e.name e_name,
  d.name d_name,
  e.salary
from employee e
inner join department d
  on e.deptid = d.id
where e.salary in (select max(salary)
                   from employee
                   group by deptid);

或使用子查询：

select e1.name e_name,
  d.name d_name,
  e1.salary
from employee e1
inner join
(
  select max(salary) salary, deptid
  from employee
  group by deptid
) e2
  on e1.salary = e2.salary
  and e1.deptid = e2.deptid
inner join department d
  on e1.deptid = d.id

参见SQL Fiddle with Demo

现在，MySQL 允许您应用聚合函数，而不是将 a 应用GROUP BY到选择列表中的非聚合字段（这不能在 sql server、oracle 等中完成）。所以你可以用来得到相同的结果：

select e.name e_name,
  d.name d_name,
  max(e.salary) salary
from employee e
inner join department d
  on e.deptid = d.id
group by d.name

参见SQL fiddle with Demo

Answer 2

Table "emp" has 
id, name,d_id,salary
and Table "department" has 
id, dname

字段。

下面的查询将输出带有部门名称的最高薪水

    SELECT  E.id,
            E.name,
            D.dname,
            max(E.salary) as higest_salary 
    FROM `emp` as E 
            left join department as D 
    on D.id=E.d_id 
    group by E.d_id

Answer 3

此查询为您提供部门列表，如果存在，则具有该部门的最高薪水，否则为 null。在这种情况下，选择员工姓名不会为您提供正确的姓名，而只会返回链接部门中的第一个员工！

SELECT 
  d.name, 
  MAX(e.salary)
FROM 
  department d 
LEFT OUTER JOIN employee e ON (e.department_id = d.id)
GROUP BY d.id

参见SQL Fiddle

如果您想要一份薪水最高的部门列表和员工姓名：

SELECT 
  d.name, 
  e.name, e.salary
FROM 
  department d 
LEFT OUTER JOIN employee e ON (e.department_id = d.id)
WHERE e.salary IN (
  SELECT MAX(em.salary) FROM employee em 
  WHERE em.department_id = d.id
);

参见SQL Fiddle

Answer 4

对于 SQL Server 2008 不是最佳解决方案……但完全可以处理从 Oracle 10G 迁移的 HR 数据库

select e.DEPARTMENT_ID,d.MaxSalary,es.FIRST_NAME,dm.MinSalary,esd.FIRST_NAME
from EMPLOYEES e
 join (select department_id,MAX(salary) MaxSalary from EMPLOYEES group by DEPARTMENT_ID) d
on e.DEPARTMENT_ID=d.DEPARTMENT_ID
 join (select first_name,DEPARTMENT_ID from EMPLOYEES ess where SALARY in (select MAX(salary) from EMPLOYEES where DEPARTMENT_ID=ess.DEPARTMENT_ID)) es
on e.DEPARTMENT_ID=es.DEPARTMENT_ID
 join (select department_id,min(salary) MinSalary from EMPLOYEES group by DEPARTMENT_ID) dm
on e.DEPARTMENT_ID=dm.DEPARTMENT_ID
 join (select first_name,DEPARTMENT_ID from EMPLOYEES ess where SALARY in (select min(salary) from EMPLOYEES where DEPARTMENT_ID=ess.DEPARTMENT_ID )) esd
on e.DEPARTMENT_ID=esd.DEPARTMENT_ID
group by e.DEPARTMENT_ID,d.MaxSalary,es.FIRST_NAME,dm.MinSalary,esd.FIRST_NAME

Answer 5

从分组中选择empname,MAX( salary)employeedep_id

以上查询将生成准确的结果。

Answer 6

从emp中选择ename，其中薪水在（按部门从emp组中选择max（salary））；