我记录了用户观看一系列视频的次数。现在我正在试图制作每天观看任何视频的用户数量。导轨COUNT SELECT DISTINCT
UserVideoWatching.where("created_at >= ? AND user_id != ?",1.month.ago, User.elephant.id).group("DATE(created_at)").reorder('created_at').count
产生SQL
SELECT COUNT(*) AS count_all, DATE(created_at) AS date_created_at FROM `user_video_watchings` WHERE (created_at >= '2013-01-27 10:43:24' AND user_id != 7) GROUP BY DATE(created_at) ORDER BY created_at
产生的每一天的所有视频观看正确的结果,但我说我想只显示每个用户一次。
我想这些SQL
SELECT COUNT(DISTINCT user_id) AS count_all, DATE(created_at) AS date_created FROM `user_video_watchings` WHERE (created_at >= '2013-01-27 10:33:18' AND user_id != 7) GROUP BY DATE(created_at) ORDER BY created_at
,所以我想
UserVideoWatching.where("created_at >= ? AND user_id != ?",1.month.ago, User.elephant.id).group("DATE(created_at)").reorder('created_at').select('COUNT(DISTINCT user_id) AS count_all, DATE(created_at) AS date_created')
会做什么,我想要的。但是这给出了
[#<UserVideoWatching >, #<UserVideoWatching >]
而不是散列。
任何想法?
我使用轨道3.1和MySQL
这是更好的方式 - 感谢 – Edward 2013-02-27 15:17:02
在轨道4,5看到http://stackoverflow.com/a/弃用19362288/670433为使用ModelName.distinct.count(:attribute_name)的解决方案 – s2t2 2014-02-07 22:39:39
如何返回模型的所有属性? – 2014-10-09 18:20:40