如何獲取模型實例及其包含，以及包含'includes？

Question

目標：用戶登錄，並看到他們訂閱的電視節目的下一集節目列表。因此，有三個常規表（用戶，TvShows，Episodes）和一個連接表（訂閱，即加入用戶和TvShow）。

的車型：

• TvShow
  • has_many :users, through: :subscriptions
  • has_many :episodes
  • has_many :subscriptions
• 情節，belongs_to :tv_show
• 用戶
  • has_many :tv_shows, through: :subscriptions
  • has_many :subscriptions
• 認購
  • belongs_to :tv_show
  • belongs_to :user

要獲取用戶訂閱的節目列表，我可以說current_user.subscriptions(includes: :tv_show).all。我想要選擇當前用戶的訂閱，包括電視節目，並且包括該節目的內容。

如果可能的話，甚至更具體，我想只包括一個插曲：以儘可能低的airtime值插曲（通話時間是劃時代的時間）> Time.now.to_i（即該播出下一集）。

我已經試過在文檔和Google上搜索這個，但是對於我來說這是一個很難用30個字來描述的問題，所以我沒有太多的運氣來解決它。我想在這裏得到一些幫助。

Answer 1

添加HAS_ONE關係，最新一集：

class TvShow 
    has_many :episodes 
    has_one :latest_episode, -> { order(:airtime) }, class_name: 'Episode', 
end 

---

User.eager_load(tv_shows: :latest_episode).find(1) 

---

查詢1

SELECT DISTINCT "users"."id" FROM "users" 
LEFT OUTER JOIN "subscriptions" ON "subscriptions"."user_id" = "users"."id" 
LEFT OUTER JOIN "tv_shows" ON "tv_shows"."id" = "subscriptions"."tv_show_id" 
LEFT OUTER JOIN "episodes" ON "episodes"."tv_show_id" = "tv_shows"."id" 
WHERE "users"."id" = ? LIMIT 1 [["id", 1]] 

查詢2

SELECT 
    "users"."id" AS t0_r0, 
    "users"."created_at" AS t0_r1, 
    "users"."updated_at" AS t0_r2, 
    "tv_shows"."id" AS t1_r0, 
    "tv_shows"."created_at" AS t1_r1, 
    "tv_shows"."updated_at" AS t1_r2, 
    "episodes"."id" AS t2_r0, 
    "episodes"."tv_show_id" AS t2_r1, 
    "episodes"."airtime" AS t2_r2, 
    "episodes"."created_at" AS t2_r3, 
    "episodes"."updated_at" AS t2_r4 
FROM "users" 
    LEFT OUTER JOIN "subscriptions" 
     ON "subscriptions"."user_id" = "users"."id" 
    LEFT OUTER JOIN "tv_shows" 
     ON "tv_shows"."id" = "subscriptions"."tv_show_id" 
    LEFT OUTER JOIN "episodes" 
     ON "episodes"."tv_show_id" = "tv_shows"."id" 
WHERE "users"."id" = ? AND "users"."id" IN (1) [["id", 1]] 

Answer 2

試試下面的代碼

current_user.subscriptions.includes({tv_show: :episodes}).where("episodes.airtime > ?", Time.now.to_i)