选择该部门薪资最高的所有部门的所有员工

Question

在其中一次采访中，有人问我下面的问题

＆＃34;写一个查询，找出所有部门中薪水最高的所有员工，包括部门名称，员工姓名和工资＆＃34;

这意味着员工表中有100条记录，部门表中有10条记录。 因此，它需要从查询中提供10条记录，如果任何部门中没有员工，则仍需要显示该部门名称。

由于

Answer 1

如果没有看到表格结构，我会说你可以用几种不同的方式做到这一点。

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

参见两者的SQL Fiddle with Demo

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

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

请参阅SQL fiddle with Demo

Answer 2

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

字段。

以下查询将输出部门名称

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

此查询为您提供一个部门列表，其中包含该部门的最高工资（如果存在），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名称，只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

见SQL Fiddle

如果您希望获得薪水和员工姓名最高的部门列表：

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
);

见SQL Fiddle

Answer 4

对于SQL Server 2008 不是最佳解决方案......但是要完全使用从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

SELECT empname，MAX（salary）FROM employee GROUP BY dep_id

以上查询将生成准确的结果。

Answer 6

从emp中选择ename的工资（按部门从emp group中选择max（salary））;