部门工资前三高的所有员工 - LeetCode
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。
通用答案:
select d.Name as 'Department', e.Name as 'Employee', e.Salary as 'Salary' from Employee e join Department d on e.DepartmentId = d.Id where 3 > ( select count(distinct e2.Salary) from Employee e2 where e2.DepartmentId = e.DepartmentId and e2.Salary > e.Salary )
MySQL解答:
# Write your MySQL query statement below select d.Name as Department, temp.Name Employee, temp.Salary from Department d left join (select e.DepartmentId, e.Name, @curRank := if (@prevDept = DepartmentId, if(@prevSal = e.Salary, @curRank, @curRank + 1), 1) as Rank, @prevSal := e.Salary as Salary, @prevDept := e.DepartmentId from Employee e, (select @prevDept := null, @curRank := 0, @prevSal := null) t order by e.DepartmentId, e.Salary desc ) temp on d.Id = temp.DepartmentId where temp.Rank <= 3
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/department-top-three-salaries
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