Question

我已经阅读了有关此问题的多个问题，但尚未找到适合我情况的答案。

我有3个模型：Apps，AppsGenres和Genres

以下是各自的相关字段：

Apps
application_id

AppsGenres
genre_id
application_id

Genres
genre_id

这里的关键是我不使用这些模型中的id字段。

我需要根据application_id和genre_id字段关联表格。

这是我目前得到的，但它并没有让我得到我需要的查询：

class Genre < ActiveRecord::Base
  has_many :apps_genres, :primary_key => :application_id, :foreign_key => :application_id
  has_many :apps, :through => :apps_genres
end

class AppsGenre < ActiveRecord::Base
  belongs_to :app, :foreign_key => :application_id
  belongs_to :genre, :foreign_key => :application_id, :primary_key => :application_id
end

class App < ActiveRecord::Base
  has_many :apps_genres, :foreign_key => :application_id, :primary_key => :application_id
  has_many :genres, :through => :apps_genres
end

供参考，这是我最终需要的查询：

@apps = Genre.find_by_genre_id(6000).apps

SELECT "apps".* FROM "apps" 
   INNER JOIN "apps_genres" 
      ON "apps"."application_id" = "apps_genres"."application_id" 
   WHERE "apps_genres"."genre_id" = 6000

Answer 1

更新试试这个：

class App < ActiveRecord::Base
  has_many :apps_genres, :foreign_key => :application_id
  has_many :genres, :through => :apps_genres
end

class AppsGenre < ActiveRecord::Base
  belongs_to :genre, :foreign_key => :genre_id, :primary_key => :genre_id
  belongs_to :app, :foreign_key => :application_id, :primary_key => :application_id
end

class Genre < ActiveRecord::Base
  has_many :apps_genres, :foreign_key => :genre_id
  has_many :apps, :through => :apps_genres
end

使用查询：

App.find(1).genres

它生成：

SELECT `genres`.* FROM `genres` INNER JOIN `apps_genres` ON `genres`.`genre_id` = `apps_genres`.`genre_id` WHERE `apps_genres`.`application_id` = 1

并查询：

Genre.find(1).apps

产生

SELECT `apps`.* FROM `apps` INNER JOIN `apps_genres` ON `apps`.`application_id` = `apps_genres`.`application_id` WHERE `apps_genres`.`genre_id` = 1

has_many：通过外键？

1 个答案: