mysql - MySQL 按期间从 Orders 中计算不同的客户

Question

我有这个东西

mysql> explain Order;
+------------------------+-------------------+------+-----+-----------+----------------+
| Field                  | Type              | Null | Key | Default   | Extra          |
+------------------------+-------------------+------+-----+-----------+----------------+
| id                     | int(11) unsigned  | NO   | PRI | NULL      | auto_increment |
| date                   | timestamp         | NO   |     | NULL      |                |
| customer               | int(11) unsigned  | NO   |     | NULL      |                |
| address                | int(11) unsigned  | NO   |     | NULL      |                |
+------------------------+-------------------+------+-----+-----------+----------------+

我需要在一年中逐月统计所有活跃客户，例如：

SELECT 
  DATE_FORMAT(`date`,'%m/%Y') as period,
  COUNT(DISTINCT(customer)) as total

FROM
  Order

WHERE
  YEAR(`date`) = '2012'

GROUP BY
  period

但是带有 DISTINCT 的 GROUP BY 效果不佳，并且此 SQL 在同一时期返回了很多结果

@edit会导致这个

07/2012 1
07/2012 1
06/2012 1
09/2012 1
12/2012 769
06/2012 1
07/2012 1
07/2012 1
06/2012 1
06/2012 1
10/2012 1
... a lot of results with 1 as total

我期待这个

01/2012 329
02/2012 279
03/2012 229
04/2012 379
05/2012 411
06/2012 152
07/2012 277
08/2012 411
09/2012 468
10/2012 501
11/2012 488
12/2012 593

Answer 1

如果日期是日期时间或时间戳，您当前的查询应该可以工作：

SELECT DATE_FORMAT(`date`,'%m/%Y') as period,
  COUNT(distinct customer) as total
FROM Orders
WHERE YEAR(`date`) = 2012
GROUP BY period

看演示