3

我有以下型号:

class Product < ActiveRecord::Base
  has_many :product_recommendation_sets, :dependent => :destroy
  has_many :recommendation_sets, :through => :product_recommendation_sets
end

class ProductRecommendationSet < ActiveRecord::Base
  belongs_to :product
  belongs_to :recommendation_set
end

class RecommendationSet < ActiveRecord::Base  
  has_many :product_recommendation_sets, :dependent => :destroy
  has_many :products, :through => :product_recommendation_sets

  has_many :recommendation_recommendation_sets, :dependent => :destroy
  has_many :recommendations, :through => :recommendation_recommendation_sets
end

class RecommendationRecommendationSet < ActiveRecord::Base
  belongs_to :recommendation
  belongs_to :recommendation_set
end

class Recommendation < ActiveRecord::Base
   has_many :recommendation_recommendation_sets, :dependent => :destroy
   has_many :recommendations, :through => :recommendation_recommendation_sets
end

我试图通过执行以下操作选择所有recommendationsproduct_id = x

RecommendationSet.joins(:products, :recommendations).where(product_id:1)

但是我得到一个未知的列错误。如何通过给定的 product_id 加入选择所有推荐。伪代码:查找recommendation_sets位置product_id = ?。找到recommendations在哪里recommendation_set_id = ?

4

1 回答 1

11

你非常接近,在你的情况下这应该工作:

RecommendationSet.joins(:products, :recommendations).where(products: { id: 1 })

请记住,在 where 子句中,您必须在条件散列中使用表的名称,而不是关系的名称。

例如,考虑以下关系:

User belongs_to :group
Group has_many :users

注意语法(复数/单数):

User.joins(:group).where(groups: { name: 'Admin' })
#               ^             ^
Group.joins(:users).where(users: { id: 15 })
#                ^            ^
于 2013-09-24T21:09:53.247 回答