Question

目标：用户登录，并查看他们订阅的电视节目的下一集节目列表。所以有三个常规表（Users，TvShows，Episodes）和一个连接表（Subscriptions，它们连接User和TvShow）。

模特：

TvShow
- has_many :users, through: :subscriptions
- has_many :episodes
- has_many :subscriptions
插曲，belongs_to :tv_show
用户
- has_many :tv_shows, through: :subscriptions
- has_many :subscriptions
订阅
- belongs_to :tv_show
- belongs_to :user

要获取用户订阅的节目列表，我可以说current_user.subscriptions(includes: :tv_show).all。

这里是我陷入困境的地方：我想选择当前用户的订阅，包括电视节目，以及包括该节目的剧集。

如果可能，更具体地说，我只想包含一个剧集：可能airtime值最低的剧集（播出时间为纪元时间）> Time.now.to_i （即下一集播出的剧集）。

我已尝试在文档和Google上搜索此内容，但对于我来说，用30个字或更少的字描述这是一个很难的问题，所以我没有太多运气来解决它。我会在这里得到一些帮助。

Answer 1

尝试以下代码

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

Answer 2

将has_one关系添加到最新一集：

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

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

查询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]]

查询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]]

如何获取模型实例及其包含，以及包含'includes？

2 个答案: