我有一个包含以下各列的销售表：
| Customer_Id | amount | date |
最好的方法是按
customer_id分组数据，并在不同的月份列上显示每个Customer_id的每月总金额（SUM）（每个Customer_id一行）？所需的输出如下所示：
Customer |January | February | March | ....
Customer_id |SUM amount | SUM amount | SUM amount | ....
我相信在Sql中这称为数据透视表。
谢谢！
最佳答案
假设您有下表：
mysql> select * from sales;
+-------------+--------+------------+
| customer_id | amount | date |
+-------------+--------+------------+
| 1 | 12 | 2015-01-01 |
| 1 | 1 | 2015-01-02 |
| 1 | 663 | 2015-02-12 |
| 2 | 22 | 2015-01-03 |
| 2 | 21 | 2015-02-12 |
| 2 | 11 | 2015-02-12 |
| 2 | 9 | 2015-04-12 |
+-------------+--------+------------+
您可以使用以下查询执行此操作：
SELECT
customer_id,
sum(if(month(date) = 1, amount, 0)) AS Jan,
sum(if(month(date) = 2, amount, 0)) AS Feb,
sum(if(month(date) = 3, amount, 0)) AS Mar,
sum(if(month(date) = 4, amount, 0)) AS Apr,
sum(if(month(date) = 5, amount, 0)) AS May,
sum(if(month(date) = 6, amount, 0)) AS Jun,
sum(if(month(date) = 7, amount, 0)) AS Jul,
sum(if(month(date) = 8, amount, 0)) AS Aug,
sum(if(month(date) = 9, amount, 0)) AS Sep,
sum(if(month(date) = 10, amount, 0)) AS Oct,
sum(if(month(date) = 11, amount, 0)) AS Nov,
sum(if(month(date) = 12, amount, 0)) AS `Dec`
FROM sales
GROUP BY customer_id;
并输出：
+-------------+------+------+------+------+------+------+------+------+------+------+------+------+
| customer_id | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
+-------------+------+------+------+------+------+------+------+------+------+------+------+------+
| 1 | 13 | 663 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 2 | 22 | 32 | 0 | 9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+-------------+------+------+------+------+------+------+------+------+------+------+------+------+
关于mysql - 按客户分组并按月列透视查询显示的MySQL销售表，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/28734019/