我的表格如下:
Tweet(id: integer, from_user: string, tweet_text: string, tweet_id: integer, course_id: integer, tweet_already_exists: boolean, tweet_posted_to_reviews: boolean, created_at: datetime, updated_at: datetime)
并且,就特定课程而言,我要获得最后50条信,用户是独一无二的。 :
Tweet.order("tweet_id desc").find_all_by_course_id(@course.id).first(50)
问题在于,如果用户不止一次地盘问,他们似乎有两倍(两条信条都出现在上面),我只想把此人带上最新的线。
Thank you!
它是一个很复杂的问题,但我不敢肯定,你想以更简单的方式来做:
Tweet.where("tweets.updated_at >= (SELECT MAX(t.updated_at) FROM tweets AS t WHERE t.from_user=tweets.from_user)").order("tweets.tweet_id desc").find_all_by_course_id(@course.id).first(50)
我先对它进行了测试,但尝试:
Tweet.find(:all, :select => DISTINCT from_user , :order => tweet_id desc )
将向你们提供一张没有重复用户的ets子。 学习这些课程的结果。
晚上,你还想使用瓦解方法uniq,使一个阵列无重复。
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