has_many:通过外键? [英] has_many :through with a foreign key?

问题描述

我已阅读有关此问题的多个问题，但尚未找到适合我情况的答案.

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

我有 3 个模型:Apps、AppsGenres 和 Genres

I have 3 models: Apps, AppsGenres and Genres

以下是每个字段的相关字段:

Here are the pertinent fields from each of those:

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.

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

我需要根据那些 application_id 和 genre_id 字段关联表格.

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

这是我目前得到的，但没有得到我需要的查询:

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

作为参考，这是我最终需要的查询:

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

推荐答案

已更新试试这个:

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

有查询:

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

生成:

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

这篇关于has_many:通过外键?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！