Ruby-on-rails has_many :通过外键？
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has_many :through with a foreign key?
提问by Shpigford
I've read multiple questions about this, but have yet to find an answer that works for my situation.
我已经阅读了多个关于此的问题，但还没有找到适合我的情况的答案。
I have 3 models: Apps, AppsGenresand Genres
我有 3 个模型：Apps,AppsGenres和Genres
Here are the pertinent fields from each of those:
以下是每个领域的相关领域：
Apps
application_id
AppsGenres
genre_id
application_id
Genres
genre_id
The key here is that I'm notusing the idfield from those models.
这里的关键是我没有使用id这些模型中的字段。
I need to associate the tables based on those application_idand genre_idfields.
我需要根据这些application_id和genre_id字段关联表。
Here's what I've currently got, but it's not getting me the query I need:
这是我目前所拥有的，但它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
For reference, here is the query I ultimately need:
作为参考，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
回答by Anton Grigoryev
UPDATEDTry this:
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
With query:
随着查询：
App.find(1).genres
It generates:
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
And query:
并查询：
Genre.find(1).apps
generates:
产生：
SELECT `apps`.* FROM `apps` INNER JOIN `apps_genres` ON `apps`.`application_id` = `apps_genres`.`application_id` WHERE `apps_genres`.`genre_id` = 1
