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[184_MEDIUM] Department Highest Salary 다시풀기 본문

SQL/LEETCODE

[184_MEDIUM] Department Highest Salary 다시풀기

소금깨 2022. 9. 26. 14:55

Table: Employee

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
id is the primary key column for this table.
departmentId is a foreign key of the ID from the Department table.
Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.

 

Table: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id is the primary key column for this table.
Each row of this table indicates the ID of a department and its name.

 

Write an SQL query to find employees who have the highest salary in each of the departments.

Return the result table in any order.

The query result format is in the following example.

 

Example 1:

Input: 
Employee table:
+----+-------+--------+--------------+
| 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 table:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
Output: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
| IT         | Max      | 90000  |
+------------+----------+--------+
Explanation: Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.

 

문제 조건

해당 부서에서 salary를 가장 많이 받는 사람을 출력하시오 

 

쿼리

with temp as(
    select e.name as Employee,
            e.salary as Salary,
            d.name as Department,
            rank() over(partition by departmentId order by salary desc) as rnk
    from employee e
        join department d on e.departmentId = d.id)

select Department,
        Employee,
        Salary
from temp
where rnk = 1

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