LeetCode数据库---1179. 重新格式化部门表

2022.05.01.劳动节快乐！

题目：

1179. 重新格式化部门表

部门表 Department：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| revenue       | int     |
| month         | varchar |
+---------------+---------+
(id, month) 是表的联合主键。
这个表格有关于每个部门每月收入的信息。
月份（month）可以取下列值 ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]。
 

编写一个 SQL 查询来重新格式化表，使得新的表中有一个部门 id 列和一些对应 每个月 的收入（revenue）列。

查询结果格式如下面的示例所示：

Department 表：
+------+---------+-------+
| id   | revenue | month |
+------+---------+-------+
| 1    | 8000    | Jan   |
| 2    | 9000    | Jan   |
| 3    | 10000   | Feb   |
| 1    | 7000    | Feb   |
| 1    | 6000    | Mar   |
+------+---------+-------+

查询得到的结果表：
+------+-------------+-------------+-------------+-----+-------------+
| id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+------+-------------+-------------+-------------+-----+-------------+
| 1    | 8000        | 7000        | 6000        | ... | null        |
| 2    | 9000        | null        | null        | ... | null        |
| 3    | null        | 10000       | null        | ... | null        |
+------+-------------+-------------+-------------+-----+-------------+

注意，结果表有 13 列 (1个部门 id 列 + 12个月份的收入列)。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reformat-department-table

=+=+=+

学到：行转列

知识点：

1.  分组： group by

2. 筛选月份，每个月份是一列：case [when...then...] end 取当前月份

3. 用聚合函数：sum() 。如果没有使用聚合函数，筛选出来的是group by, case...end之后的第一行

1. 第一步，用group by id 让表Department按 id 分组
Department 表
+------+---------+-------+
| id   | revenue | month |
+------+---------+-------+
| 1    | 8000    | Jan   |
| 2    | 9000    | Jan   |
| 3    | 10000   | Feb   |
| 1    | 7000    | Feb   |
| 1    | 6000    | Mar   |
+------+---------+-------+

group by id, 数据被分为三组，如下：
+------+---------+-------+
| id   | revenue | month |
+------+---------+-------+
| 1    | 8000    | Jan   |
| 1    | 7000    | Feb   |
| 1    | 6000    | Mar   |
-------------------------
| 2    | 9000    | Jan   |
-------------------------
| 3    | 10000   | Feb   |
+------+---------+-------+

2. 第二步，用case [when...then...] end
   但没有聚合函数 当一个单元格中有多个数据时，只输出第一行数据
　　如：当id=1时, (case when month='Feb' then revenue end)= [NULL, 7000, NULL], 
   没有聚合函数会只取第一个，即NULL

3.第三部，使用聚和函数
  解决单元格内多个数据的情况。
  sum(case when month='Feb' then revenue end),当id=1时,
  从Jan、Feb、Mar中找到了符合条件的Feb，并最终返回对应的revenue的值，即7000

 sql:

select id,
    sum(case month when 'Jan' then revenue end) as Jan_Revenue,
    sum(case month when 'Feb' then revenue end) as Feb_Revenue,
    sum(case month when 'Mar' then revenue end) as Mar_Revenue,
    sum(case month when 'Apr' then revenue end) as Apr_Revenue,
    sum(case month when 'May' then revenue end) as May_Revenue,
    sum(case month when 'Jun' then revenue end) as Jun_Revenue,
    sum(case month when 'Jul' then revenue end) as Jul_Revenue,
    sum(case month when 'Aug' then revenue end) as Aug_Revenue,
    sum(case month when 'Sep' then revenue end) as Sep_Revenue,
    sum(case month when 'Oct' then revenue end) as Oct_Revenue,
    sum(case month when 'Nov' then revenue end) as Nov_Revenue,
    sum(case month when 'Dec' then revenue end) as Dec_Revenue
from Department
group by id
order by id;