的has_many：通过与外键？

问题描述

我读过有关这多个问题，但还没有找到适合我的情况问题的解答。

我有3个型号：应用， AppsGenres 和流派

下面是相关领域从每次的：

 应用程序
APPLICATION_IDAppsGenres
genre_id
APPLICATION_ID流派
genre_id
 

这里的关键是从这些模型，我的不可以使用 ID 字段。

我需要根据这些表关联 APPLICATION_ID 和 genre_id 字段。

下面就是我目前得到的，但它没有得到我，我需要查询：

 类体裁＆LT; ActiveRecord的::基地
  的has_many：apps_genres，：primary_key =＆GT; ：APPLICATION_ID，：foreign_key =＆GT; ：APPLICATION_ID
  的has_many：应用程序，：通过=＆GT; ：apps_genres
结束类AppsGenre＆LT; ActiveRecord的::基地
  belongs_to的：应用程序，：foreign_key =＆GT; ：APPLICATION_ID
  belongs_to的：体裁，：foreign_key =＆GT; ：APPLICATION_ID，：primary_key =＆GT; ：APPLICATION_ID
结束类应用＆LT; ActiveRecord的::基地
  的has_many：apps_genres，：foreign_key =＆GT; ：APPLICATION_ID，：primary_key =＆GT; ：APPLICATION_ID
  的has_many：流派：通过=＆GT; ：apps_genres
结束
 

有关参考，在这里是查询我最终需要：

  @apps = Genre.find_by_genre_id（6000）.apps选择应用程序* FROM应用程序
   INNER JOINapps_genres
      ON应用程序。APPLICATION_ID=apps_genres。APPLICATION_ID
   WHEREapps_genres。genre_id= 6000
 

解决方案

更新时间：试试这个：

 类应用＆LT; ActiveRecord的::基地
  的has_many：apps_genres，：foreign_key =＆GT; ：APPLICATION_ID
  的has_many：流派：通过=＆GT; ：apps_genres
结束类AppsGenre＆LT; ActiveRecord的::基地
  belongs_to的：体裁，：foreign_key =＆GT; ：genre_id，：primary_key =＆GT; ：genre_id
  belongs_to的：应用程序，：foreign_key =＆GT; ：APPLICATION_ID，：primary_key =＆GT; ：APPLICATION_ID
结束类体裁＆LT; ActiveRecord的::基地
  的has_many：apps_genres，：foreign_key =＆GT; ：genre_id
  的has_many：应用程序，：通过=＆GT; ：apps_genres
结束
 

通过查询：

  App.find（1）.genres
 

它产生：

  SELECT`genres`。* FROM`genres` INNER JOIN`apps_genres`开`genres`.`genre_id` =`apps_genres`.`genre_id`其中`apps_genres`.`application_id `= 1
 

和查询：

  Genre.find（1）.apps
 

生成：

  SELECT`apps`。* FROM`apps` INNER JOIN`apps_genres`开`apps`.`application_id` =`apps_genres`.`application_id`其中`apps_genres`.`genre_id `= 1
 

I've read multiple questions about this, but have yet to find an answer that works for my situation.

I have 3 models: Apps, AppsGenres and Genres

Here are the pertinent fields from each of those:

Apps
application_id

AppsGenres
genre_id
application_id

Genres
genre_id

The key here is that I'm not using the id field from those models.

I need to associate the tables based on those application_id and genre_id fields.

Here's what I've currently got, but it's not getting me the query I need:

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

For reference, here is the query I ultimately need:

@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

解决方案

UPDATED Try this:

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

With query:

App.find(1).genres

It generates:

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

And query:

Genre.find(1).apps

generates:

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

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