ActiveRecord查询，连接(Rails 3.1)

Question

如何使用

获取JSON对象?

[
  {
    Property.field_1,
    Property.field_n,
    PropAssignmConsumer.field_1,
    PropAssignmConsumer.field_n
  },
  {
    Property.field_1,
    Property.field_n,
    PropAssignmConsumer.field_1,
    PropAssignmConsumer.field_n
  },
  ...,
  {
    Property.field_1,
    Property.field_n,
    PropAssignmConsumer.field_1,
    PropAssignmConsumer.field_n
  }
]

排序的一些关键(可以在属性或PropAssignmConsumer字段)为给定的user_entity对象?例如，获取链接到给定消费者/user_entity的所有属性，从properties和prop_assignm_consumers中提取字段，按properties或prop_assignm_consumer表中的字段排序。

这些是我的模型:

class Property < ActiveRecord::Base
  has_many    :prop_assignm_consumers,  :dependent => :restrict
end
class PropAssignmConsumer < ActiveRecord::Base
  belongs_to  :consumer
  belongs_to  :property
end
class Consumer < UserEntity
  has_many        :prop_assignm_consumers,    :dependent => :destroy
  has_many        :properties,                :through   => :prop_assignm_consumers
end

我正在做

properties = user_entity.properties.find(:all, :order => "#{sort_key} #{sort_ord}")
properties.each do |p|
  a = p.prop_assignm_consumers.find_by_consumer_id(current_user.user_entity.id)
  ... do something with a and p....
end

但这似乎效率不高....

Answer 1

也许我错过了什么。为什么你的财产没有提到消费者?你有很多对很多，你只是没有完成它。只要加上has_many :consumers, :through => :prop_assignm_consumer就可以做

properties = user_entity_properties.all(:include => :consumers)
properties.each do |p| 
  p.consumers.where(:id => current_user.user_entity.id)
end 

虽然现在我们写了这个，并且考虑到你做的是find_by而不是find_all_by，很明显只有1。所以你可以走另一条路。

consumer = Consumer.where(:id => current_user.user_entity.id).includes(:properties).first
consumer.properties.each do |p| 
  ... do something with p and consumer
end 

裁判http://api.rubyonrails.org/classes/ActiveRecord/Associations/ClassMethods.html