1
日
唯一登录我(为例)一个MySQL表“登录”与此内容:MySQL回报:计算每个
user_id | last_login
1 | 2015-02-01 05:01:07
1 | 2015-02-01 12:42:09
2 | 2015-02-01 22:16:23
2 | 2015-02-02 15:45:23
2 | 2015-02-04 21:27:04
3 | 2015-02-04 06:25:45
4 | 2015-02-05 03:12:01
我的问题是:我怎么能生成占所有用户的独特的夏日天。所以我会像这样的报告:
day | count
2015-02-01 | 2
2015-02-02 | 1
2015-02-04 | 2
2014-02-05 | 1
这查询不工作:
SELECT DATE_FORMAT(login_date, '%Y-%m-%d') AS `day`
(SELECT COUNT(DISTINCT(user_id)) AS count, DATE_FORMAT(login_date, '%Y-%m-%d') AS `sdate` WHERE `sdate` = day)
FROM `logins`
GROUP BY `day`
ORDER BY `day` // query not working.....
我怎样才能每天都独立用户的SQL夏日...?