Rails嵌套has_many关联，如何获得所有孩子的最后5个？

Question

假设

Post has_many :comments 

和

Comment has_many :ratings 

我怎样才能抓住每个帖子的最后5个评论收视率？我一直在考虑循环评论每篇文章，但这并不能解决最后5部分。

编辑：为响应J.，因为我似乎无法格式化注释字段

里面的代码你可以嵌套：通过关系？说...

class Category < ActiveRecord::Base 
    has_many :posts 
    has_many :comments, :through => posts 
    has_many :ratings, :through => comments 
end 

class Post < ActiveRecord::Base 
    belongs_to :category 
    has_many :comments 
    has_many :ratings, :through => comments 
end 

class Comment < ActiveRecord::Base 
    belongs_to :post 
    has_many :ratings 
end 

class Rating < ActiveRecord::Base 
    belongs_to :comment 
end 

Answer 1

最后我能得到什么，我用这个寻找下的Rails 3

class Category < ActiveRecord::Base 
    has_many :posts 
    has_many :comments, :through => :posts 

    def ratings 
    Rating.category_ratings(self) 
    end 
end 

class Rating < ActiveRecord::Base 
    belongs_to :comment 

    scope :category_ratings, lambda { |c| 
    joins(:comment, 'INNER JOIN `posts` ON `posts`.`id` = `comments`.`post_id`'). 
    where(:posts => {:category_id => c.id}). 
    select('DISTINCT `comments`.*') 
    } 
end 

Answer 2

我相信像下面这样可能工作...让我知道，如果它不：]

class Post < ActiveRecord::Base 
    has_many :comments 
    has_many :ratings, :through => :comments 
end 

class Comment < ActiveRecord::Base 
    belongs_to :post 
    has_many :ratings 
end 

class Rating < ActiveRecord::Base 
    # I'm assuming you have the created_at column 
    default_scope :order => 'created_at DESC' 
end 

# controller 
@last_five_ratings = @post.ratings.all(:limit => 5) 

Answer 3

您可以使用标准的ActiveRecord做到这一点：find :all, :order => "created_at desc", :limit => 5。我想，你可以用这个像这样一个named_scope：

class Rating < ActiveRecord::Base 
    named_scope :most_recent, lambda { |n| { :conditions => [:order => 'created_at desc', 
              :limit => n] } 

end 

and in your controller: 
@recent_ratings = @comment.ratings.most_recent(5)