Question

我有模特用户

class User < ActiveRecord::Base  
  has_many :ratings  
  has_many :rated_films, :through => :ratings, :source => :film  
end

和电影

class Film < ActiveRecord::Base  
  has_many :users, :through => :ratings  
end

我希望找到所有未被指定用户评分的电影，如

class Film < ActiveRecord::Base  
  has_many :users, :through => :ratings  
  named_scope :not_rated_by_user, lambda { |user|  
    {:joins => :users, :conditions => ['? NOT IN users', user]}  
  }  
end  

Film.not_rated_by_user(User.first)

我不熟悉SQL，所以我不太确定是否可以在命名范围内实现。

非常感谢

尤里

Answer 1

我想你有一个ratings表，这是你的join表。对？所以你需要这样的东西：

class User < ActiveRecord::Base  
   has_many :ratings
   has_many :rated_films, :through => :ratings, :source => :film  
end

class Film < ActiveRecord::Base  
   has_many :ratings
   has_many :users, :through => :ratings

   named_scope :not_rated_by_user, lambda { |user_id| {
      :include => :ratings,
      :conditions => ['? NOT IN (ratings.user_id)', user_id]
   }}
end

class Rating < ActiveRecord::Base
   belongs_to :film
   belongs_to :user
end

你可以使用

Film.not_rated_by_user(User.first.id)

如果有帮助，请告诉我。我没有测试过！

named_scope和HABTM关联

1 个答案: