【MySQL】每個類別中取前三高

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2023-04-10 15:06:36 By 過路君子

大家好,這裡是快忘光SQL的小編過路君子

看了看紀錄,上次寫這麼複雜的SQL已經是在一年之前

 

 

SQL 小編說常用也常用,說不常用也不常用,為什麼這麼說呢?

因為平常使用的時候就只會用基本的 SELECT 功能,什麼 sub select 根本就不會用到。

今天回來複習一下 MySQL 的 SQL 語法

 

那在開始之前,先假設我們有以下的資料表:

Table: Employee
+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+

Table: Department
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+

並且具有以下的內容:

Employee table:
+----+-------+--------+--------------+
| 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 table:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+

 

不考慮資料重複或並列

我們則需要輸出以下的資料:

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

簡單來說就是將 departmentId 轉換成對應的部門名稱,然後要將每個前三高薪水(Salary)的人列出,最後在對資料做排序。

 

直接上 SQL 語法:

SELECT c.name AS Department, a.name AS Employee, a.salary AS Salary
FROM Employee AS a
LEFT JOIN Department AS c
  ON c.id = a.departmentId
WHERE
  (
    SELECT COUNT(b.salary)
    FROM Employee AS b
    WHERE b.departmentId = a.departmentId
      AND a.salary < b.salary
  ) < 3
ORDER BY c.id, a.salary;

輸出結果就會如上方所示。

其中的 LEFT JOIN 作用就是為了要讓 Department 表裡的資料直接接在 Employee 表的後面,只有當 Department 表內的 id 和 Employee 表內的 departmentId 相同才需要做串接。

而比較重要的就是 WHERE 內的 sub select 了,此 sub select 會試著去統計在 Employee 表內目前有幾個人的薪水大於目前所執行到人的薪水。

而最後的小於三則是取前三高,如果要取前五高就改成 5 即可。

 

考慮資料重複或並列

若需考慮資料重複,那我們要輸出的資料變成如下:

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

這應該才算是比較常用到的手段,總之寫法如下:

SELECT c.name AS Department, a.name AS Employee, a.salary AS Salary
FROM Employee AS a
LEFT JOIN Department AS c
  ON c.id = a.departmentId
WHERE
  (
    SELECT COUNT(b.salary)
    FROM (SELECT DISTINCT d.departmentId, d.salary FROM Employee AS d) AS b
    WHERE b.departmentId = a.departmentId
      AND a.salary < b.salary
  ) < 3;
ORDER BY c.id, a.salary;

 

只需要改一行就可以了。

簡單來說,在開始找有幾個人薪水大於目前的人之前,先把要比較的表去除重複的資料,這樣統計的時候也不會重複統計啦。

 

 

 

後記

總之這次寫完感覺這個 SQL 很有用,所以就在這邊留存一份,省的未來在那邊推敲半天。