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按客户分组并按月列枢轴查询显示的Mysql销售表

作者:互联网

我有一个包含以下列的销售表:

|  Customer_Id | amount |  date  |

什么是按customer_id分组数据的最佳方法,并在不同的月份列上显示每个Customer_id(每个Customer_id一行)的每月总金额(SUM)?

期望的输出将是这样的:

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!

解决方法:

假设您有以下表格:

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 |
+-------------+--------+------------+

您可以使用此查询执行此操作:

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 |
+-------------+------+------+------+------+------+------+------+------+------+------+------+------+

标签:mysql,database,pivot-table
来源: https://codeday.me/bug/20191007/1867733.html