has_many: bir yabancı anahtarla mı?

Question

Bu konuyla ilgili birçok soru okudum, ancak durumum için çalışan bir yanıt bulmaya henüz başlamadım. Apps, AppsGenres ve burada Genres

olanların her birinden ilgili alanlar şunlardır:

Ben 3 modelleri var

Apps 
application_id 

AppsGenres 
genre_id 
application_id 

Genres 
genre_id 

Burada anahtar Ben değil id alanını kullanarak olduğum

bu modellerden.

Tabloları application_id ve genre_id alanlarına göre ilişkilendirmem gerekiyor. İşte

Şu anda bu var, ama bana ihtiyacım sorgu alamayacak:

@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

GÜNCELLEME: Referans olarak

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 

, burada sonuçta gerek sorgu Bu deneyin: sorgusu ile

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 

:

App.find(1).genres 

O oluşturur:

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

Ve sorguyu:

Genre.find(1).apps 

oluşturur:

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