leetcode-----mysql

185. 部门工资前三高的所有员工 - 力扣（LeetCode）

表: Employee

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
id 是该表的主键列(具有唯一值的列)。
departmentId 是 Department 表中 ID 的外键（reference 列）。
该表的每一行都表示员工的ID、姓名和工资。它还包含了他们部门的ID。

表: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id 是该表的主键列(具有唯一值的列)。
该表的每一行表示部门ID和部门名。

公司的主管们感兴趣的是公司每个部门中谁赚的钱最多。一个部门的 高收入者 是指一个员工的工资在该部门的 不同 工资中 排名前三 。

编写解决方案，找出每个部门中 收入高的员工 。

以 任意顺序 返回结果表。

返回结果格式如下所示。

示例 1:

输入: 
Employee 表:
+----+-------+--------+--------------+
| 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:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
输出: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+
解释:
在IT部门:
- Max的工资最高
- 兰迪和乔都赚取第二高的独特的薪水
- 威尔的薪水是第三高的

在销售部:
- 亨利的工资最高
- 山姆的薪水第二高
- 没有第三高的工资，因为只有两名员工

提示：

• 没有姓名、薪资和部门 完全 相同的员工。

第一步，把前面列下来

select  d.name as Department ,e.name as Employee  ,e.salary as Salary

from Department d left join Employee e

on d.id = e.departmentId

第二步：公司的主管们感兴趣的是公司每个部门中谁赚的钱最多。一个部门的 高收入者 是指一个员工的工资在该部门的 不同 工资中 排名前三 。

3 > (select count(distinct e2.salary)
           from 
                Employee e2
            where 
                e2.salary> e.salary
            and
                e.departmentId = e2.departmentId 
            )

中和

# Write your MySQL query statement below

select  d.name as Department ,e.name as Employee  ,e.salary as Salary

from Department d left join Employee e

on d.id = e.departmentId

where 3 > (select count(distinct e2.salary)

           from

                Employee e2

            where

                e2.salary> e.salary

            and

                e.departmentId = e2.departmentId

            )

分析报错，不应该用左连接的方式

# Write your MySQL query statement below

select  d.name as Department ,e.name as Employee  ,e.salary as Salary

from Department d  join Employee e

on d.id = e.departmentId

where 3 > (select count(distinct e2.salary)

           from

                Employee e2

            where

                e2.salary> e.salary

            and

                e.departmentId = e2.departmentId

            )