6

我有 3 个模型。Rom::Favorite, Rom::Card, User. 我在创建User has_many rom_cards through rom_favorites

这是我的模型的相关部分

罗::卡

class Rom::Card < ActiveRecord::Base
    has_many :rom_favorites, class_name: "Rom::Favorite", foreign_key: "rom_card_id", dependent: :destroy

    self.table_name = "rom_cards"

end

用户

class User < ActiveRecord::Base
  # Setup accessible (or protected) attributes for your model
  attr_accessible :email, :password, :password_confirmation, :remember_me, :role

  has_many :rom_favorites, class_name: "Rom::Favorite", dependent: :destroy
  has_many :rom_cards, class_name: "Rom::Card", through: :rom_favorites, class_name: "Rom::Favorite"

end

罗::最喜欢的

   class Rom::Favorite < ActiveRecord::Base
      attr_accessible :rom_card_id, :user_id

      belongs_to :user
      belongs_to :rom_card, class_name: "Rom::Card"

      validates :user, presence: true
      validates :rom_card, presence: true
      validates :rom_card_id, :uniqueness => {:scope => :user_id}

      self.table_name = "rom_favorites"

    end

关联附带的所有实用程序方法都可以工作,除了

a = User.find(1)
a.rom_cards

该调用a.rom_cards返回一个空数组,它似乎运行了这个 SQL 查询

SELECT "rom_favorites".* FROM "rom_favorites" INNER JOIN "rom_favorites" "rom_favorites_rom_cards_join" ON "rom_favorites"."id" = "rom_favorites_rom_cards_join"."rom_card_id" WHERE "rom_favorites_rom_cards_join"."user_id" = 1

我不擅长 SQL,但我认为这似乎是正确的。

我知道a.rom_cards应该返回 2 张卡片,因为a.rom_favorites返回 2 个收藏夹,并且在这些收藏夹中存在 card_id。

应该允许的调用rom_cards如下

has_many :rom_cards, class_name: "Rom::Card", through: :rom_favorites, class_name: "Rom::Favorite"

我觉得这个问题与它试图通过收藏夹查找用户卡并且它正在寻找card_id(因为我指定了类 Rom::Card)而不是 rom_card_id 的事实有关。但我可能是错的,不确定。

4

1 回答 1

14

您正在复制class_name关联哈希中的键。无需编写class_name: "Rom::Favorite",因为使用through: :rom_favorites它会使用has_many :rom_favorites.

尝试:

has_many :rom_cards, class_name: "Rom::Card", through: :rom_favorites
于 2013-09-04T19:26:20.830 回答