LeetCode：MySQL分组内取前几名问题（难度：困难）

题目：查询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 。

Department 表包含公司所有部门的信息。

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：

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。

题解

对于这种分组内取前几名的问题，可以先group by然后用having count()来筛选。思路为：
先找到每个部门工资前三的员工ID（自联结+group by+having）
再找到每个部门工资前三的部门名称、员工名称、工资和员工ID（两表联结）
最后子查询嵌套即可

1.找到每个部门工资前三的员工ID

用Employee和自己做连接，连接条件是【部门相同但工资比我高】
接下来按照having count(Salary) <= 2来筛选的原理是：如果【跟我一个部门而且工资比我高的人数】不超过2个，那么我一定是部门工资前三
这样就可以查询出每个部门工资前三的员工ID

select e1.id,e1.name,e1.DepartmentId
from employee e1 left join employee e2 
     on e1.DepartmentId=e2.DepartmentId and e1.Salary<e2.Salary
group by e1.id
having count(distinct e2.Salary)<=2;

2.外层嵌套：找到每个部门工资前三的部门名称、员工名称、工资和员工ID

select d.Name as Department,e.Name as Employee,e.Salary as Salary
from Employee as e left join Department as d 
on e.DepartmentId = d.Id
where e.Id in(#每个部门工资前三的员工ID)
and e.DepartmentId in (select Id from Department)
order by d.Id asc,e.Salary desc;

进行1和2的子查询嵌套即可得到

完整代码如下：

select d.Name as Department,e.Name as Employee,e.Salary as Salary
from Employee as e left join Department as d 
on e.DepartmentId = d.Id
where e.Id in
(
    select e1.Id
    from Employee as e1 left join Employee as e2
    on e1.DepartmentId = e2.DepartmentId and e1.Salary < e2.Salary
    group by e1.Id
    having count(distinct e2.Salary) <= 2
)
and e.DepartmentId in (select Id from Department)
order by d.Id asc,e.Salary desc