# 1179.(Easy)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>

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

## Solution

```sql
select id,
    sum(if(month="Jan",revenue,null)) Jan_Revenue,
    sum(if(month="Feb",revenue,null)) Feb_Revenue,
    sum(if(month="Mar",revenue,null)) Mar_Revenue,
    sum(if(month="Apr",revenue,null)) Apr_Revenue,
    sum(if(month="May",revenue,null)) May_Revenue,
    sum(if(month="Jun",revenue,null)) Jun_Revenue,
    sum(if(month="Jul",revenue,null)) Jul_Revenue,
    sum(if(month="Aug",revenue,null)) Aug_Revenue,
    sum(if(month="Sep",revenue,null)) Sep_Revenue,
    sum(if(month="Oct",revenue,null)) Oct_Revenue,
    sum(if(month="Nov",revenue,null)) Nov_Revenue,
    sum(if(month="Dec",revenue,null)) Dec_Revenue
from Department
group by 1
```


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