首先選擇在軌道的每一個組唯一的密鑰對的日期

Question

class User < ActiveRecord::Base 
    has_many :events 
    has_many :trainings, -> { distinct }, through: :events 
end 

class Event < ActiveRecord::Base 
    belongs_to :user 
    belongs_to :training 
end 

class Training < ActiveRecord::Base 
    has_many :events 
    has_many :user, -> { distinct }, through: :events 
end 

events table contains: 

t.integer "user_id" 
t.integer "training_id" 
t.due  "date" 

想象的事件表具有以下USER_ID，training_id和到期日

[ 1, 1, 25/06/16] 
[ 1, 1, 25/06/15] 
[ 1, 1, 25/06/14] 

[ 1, 2, 25/06/16] 
[ 1, 2, 25/06/15] 
[ 1, 2, 25/06/14] 

[ 2, 1, 25/06/16] 
[ 2, 1, 25/06/15] 
[ 2, 1, 25/06/14] 

[ 2, 2, 25/06/16] 
[ 2, 2, 25/06/15] 
[ 2, 2, 25/06/14] 

我想選擇按到期日最近的值這些「組」中的每一個（用戶&訓練密鑰對）。

即

[ 1, 1, 25/06/16] 
[ 1, 2, 25/06/16] 
[ 2, 1, 25/06/16] 
[ 2, 2, 25/06/16] 

任何想法如何實現這一目標？

Answer 1

管理使用自定義查詢。只有很小的問題是，即使它們沒有過期，它也會選擇行，只是最新的一行。如果我將<'＃{Time.now}'放入其中，那麼它基本上會選擇一個不是我想要的東西，但我可以在視圖中放入一些東西來過濾日期< Time.now作爲compramise。

@overdue = Event.find_by_sql("SELECT DISTINCT ON (user_id, training_id) 
     id, user_id, training_id, date, due FROM events 
    ORDER BY user_id, training_id, due DESC")