# 1179. Reformat Department Table (Easy)
Tags: Database
Languages available: mysql, mssql, postgresql, oraclesql, pythondata

Table: `Department`
 
```

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| revenue     | int     |
| month       | varchar |
+-------------+---------+
In SQL,(id, month) is the primary key of this table.
The table has information about the revenue of each department per month.
The month has values in ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"].

```
 
 
 
Reformat the table such that there is a department id column and a revenue column **for each month**.
 
Return the result table in **any order**.
 
The result format is in the following example.
 
 
 
**Example 1:**
 
```

**Input:** 
Department table:
+------+---------+-------+
| id   | revenue | month |
+------+---------+-------+
| 1    | 8000    | Jan   |
| 2    | 9000    | Jan   |
| 3    | 10000   | Feb   |
| 1    | 7000    | Feb   |
| 1    | 6000    | Mar   |
+------+---------+-------+
**Output:** 
+------+-------------+-------------+-------------+-----+-------------+
| id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+------+-------------+-------------+-------------+-----+-------------+
| 1    | 8000        | 7000        | 6000        | ... | null        |
| 2    | 9000        | null        | null        | ... | null        |
| 3    | null        | 10000       | null        | ... | null        |
+------+-------------+-------------+-------------+-----+-------------+
**Explanation:** The revenue from Apr to Dec is null.
Note that the result table has 13 columns (1 for the department id + 12 for the months).

```

## Starter code (mysql)
```
# Write your MySQL query statement below
```
