[英]rails COUNT SELECT DISTINCT
我正在記錄用戶觀看一系列視頻的次數。 現在,我正在嘗試制作每天觀看任何視頻的用戶數量的圖表。
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
這會為每天觀看的所有視頻產生正確的結果,但正如我所說,我只想向每個用戶展示一次。
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
所以我認為
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')
會做我想做的事。 但這給了
[#<UserVideoWatching >, #<UserVideoWatching >]
而不是哈希。
有任何想法嗎?
我正在使用rails 3.1和mysql
您可以使用distinct.count(:attribute_name)
。
(在Rails 3中使用: count(:user_id, distinct: true)
代替)
從而:
UserVideoWatching.where("created_at >= ? AND user_id != ?", 1.month.ago, User.elephant.id)
.group("DATE(created_at)").reorder('created_at').distinct.count(:user_id)
無法測試,但我認為這將產生你想要的SQL。
在Rails 4中,使用其他答案中提到的(...).uniq.count(:user_id)
(針對此問題以及SO上的其他地方)實際上會導致查詢中出現額外的DISTINCT
:
SELECT DISTINCT COUNT(DISTINCT user_id) FROM ...
我們實際需要做的是自己使用SQL字符串:
(...).count("DISTINCT user_id")
這給了我們:
SELECT COUNT(DISTINCT user_id) FROM ...
應該使用distinct,在rails 5.0.1中,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"
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