Question

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

我试图通过以下方式选择product_id = recommendations的所有x：

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

但是我收到了一个未知的列错误。如何加入选择给定product_id的所有建议。 Psudo代码：查看recommendation_sets product_id = ?。查看recommendations recommendation_set_id = ?。

Answer 1

你非常接近，在你的情况下这应该有效：

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

请记住，where子句中的必须在条件哈希中使用表名，而不是关系的名称。

作为一个例子，考虑这些关系：

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 })
#                ^            ^

Rails多个连接AR查询和where子句

1 个答案: