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。
分析题目
要求:
1. 不同部门
2. 工资前三高
解题思路
需要根据部门进行分组,分组后对数据进行排序。
但是MySql 8.0 之前无法对数据进行分组,对分组后的数据进行排序
换一种思路,同一部门,比薪资高的薪资不超过3个,那么薪资就属于前三高
那么如果比较呢,where中关联表进行对比
解题步骤
- where 中 比较,count()统计
SELECT
c.`name` AS department,
a.`name` AS employee,
a.salary
FROM
employee a
LEFT JOIN department c ON a.DepartmentId = c.id
WHERE
3 > (SELECT count(b.salary) FROM employee b WHERE a.DepartmentId = b.DepartmentId AND b.salary > a.salary)
image.png
- distinct
统计结果并不对,并不是工资前三高,而是工资高的前三个人
SELECT
c.`name` AS department,
a.`name` AS employee,
a.salary
FROM
employee a
LEFT JOIN department c ON a.DepartmentId = c.id
WHERE
3 > (SELECT count(b.salary) FROM employee b WHERE a.DepartmentId = b.DepartmentId AND b.salary > a.salary)
解法二:使用MySQL8的开窗函数 dense_rank()
select
Department,Employee,Salary
from
(
select
d.name Department,e.name Employee,e.salary Salary,dense_rank()
over(partition by d.name order by e.salary desc) rk
from
Employee e left join Department d
on e.departmentid = d.id
)t1
where rk <= 3
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