Question

我正在记录用户观看一系列视频的次数。现在，我正在尝试制作每天观看任何视频的用户数量的图表。

UserVideoWatching.where("created_at >= ? AND user_id != ?",1.month.ago, User.elephant.id).group("DATE(created_at)").reorder('created_at').count

生成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

为每天观看的所有视频产生正确的结果，但正如我所说，我只想向每个用户展示一次。

我想要的SQL是

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')

会做我想要的。但是这给了

[#<UserVideoWatching >, #<UserVideoWatching >]

而不是哈希。

有什么想法吗？

我正在使用rails 3.1和mysql

Answer 1

您可以使用distinct.count(:attribute_name)。

（在Rails 3中使用：count(:user_id, distinct: true)代替）

因此：

UserVideoWatching.where("created_at >= ? AND user_id != ?", 1.month.ago, User.elephant.id)
.group("DATE(created_at)").reorder('created_at').distinct.count(:user_id)

无法测试，但我认为这会产生你想要的SQL。

Answer 2

在Rails 4中，使用其他答案中提到的(...).uniq.count(:user_id)（针对此问题和其他地方的SO）实际上会导致查询中出现额外的DISTINCT：

SELECT DISTINCT COUNT(DISTINCT user_id) FROM ...

我们实际需要做的是自己使用SQL字符串：

(...).count("DISTINCT user_id")

这给了我们：

SELECT COUNT(DISTINCT user_id) FROM ...

Answer 3

在rails 5.0.1中应该使用distinct，不同的uniq，但是

[11] pry(main)> Needremember.distinct.count(:word)
(1.1ms)  SELECT DISTINCT COUNT(DISTINCT "needremembers"."word") FROM "needremembers"
[12] pry(main)> Needremember.uniq.count(:word)
DEPRECATION WARNING: uniq is deprecated and will be removed from Rails 5.1 (use distinct instead) (called from __pry__ at (pry):12)
   (0.6ms)  SELECT DISTINCT COUNT(DISTINCT "needremembers"."word") FROM "needremembers"

rails COUNT SELECT DISTINCT

3 个答案: