SQL练习20201202
184.部门工资最高的员工
Employee 表包含所有员工信息,每个员工有其对应的 Id, salary 和 department Id。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Jim | 90000 | 1 |
| 3 | Henry | 80000 | 2 |
| 4 | Sam | 60000 | 2 |
| 5 | Max | 90000 | 1 |
+----+-------+--------+--------------+
Department 表包含公司所有部门的信息。
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
编写一个 SQL 查询,找出每个部门工资最高的员工。对于上述表,您的 SQL 查询应返回以下行(行的顺序无关紧要)。
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Jim | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
解释:
Max 和 Jim 在 IT 部门的工资都是最高的,Henry 在销售部的工资最高。
select Department,Employee,Salary
from (
select d.Name as Department,e.Name as Employee,e.Salary as Salary,
rank() over(partition by d.Name order by e.Salary desc) as rk
from Employee e
inner join Department d
on e.DepartmentId=d.Id) a
where a.rk=1
185.部门工资前三高的所有员工
Employee 表包含所有员工信息,每个员工有其对应的工号Id,姓名Name,工资 Salary 和部门编号 DepartmentId 。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 85000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
| 7 | Will | 70000 | 1 |
+----+-------+--------+--------------+
Department 表包含公司所有部门的信息。
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
编写一个 SQL 查询,找出每个部门获得前三高工资的所有员工。例如,根据上述给定的表,查询结果应返回:
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 85000 |
| IT | Will | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+
解释:
IT 部门中,Max 获得了最高的工资,Randy 和 Joe 都拿到了第二高的工资,Will 的工资排第三。销售部门(Sales)只有两名员工,Henry 的工资最高,Sam 的工资排第二。
select Department,Employee,Salary
from(
select d.Name as Department,e.Name as Employee,e.Salary as Salary,dense_rank() over(partition by d.Id order by e.Salary desc) as dk
from Employee e
join Department d on e.DepartmentId=d.Id
) a
where dk<=3
窗口函数dense_rank():不间断的组内排序。使用这个函数时,可以出现1,1,2,2这种形式的分组。