具有多点穿越两种模式

问题描述

我在这里有两个重要的模型.还有更多与之交互的模型，但只有这些模型才有意义.

I have two models of importance here. There's more models interacting with them too, but only these should be relevant.

class Image < ActiveRecord::Base
  belongs_to :user
  belongs_to :trip
end

class Mission < ActiveRecord::Base
  belongs_to :user
  belongs_to :trip

  # has_many :images, through: :user
end

已注释掉的行将我带到一半，但我想满足图像具有与任务相同的旅程ID的其他条件.

The commented out line gets me halfway there but I want to meet an additional condition of the image having the same Trip ID as the Mission.

  # has_many :images, through: [:user, :trip]

像某些类似方法有时使用数组那样，这是无效的语法.

Using an array like some similar methods sometimes take, is invalid syntax.

   def images
     Image.where(user_id: user_id, trip_id: trip_id)
   end

我应该那样做吗?还是有更好的方法呢?我还在has_many上尝试了条件语句，但是我放置的任何动态变量都被Image::ActiveRecord_Relation

Should I just do that? Or is there a better way to do it? I also tried with conditionals on the has_many but anything dynamic I put there was being called off of Image::ActiveRecord_Relation

以下是一些示例记录:

<User id=1>
<User id=2>
<Trip id=1>
<Trip id=2>
<Image id=1, user_id=1, trip_id=1>
<Image id=2, user_id=1, trip_id=1>
<Image id=3, user_id=1, trip_id=1>
<Mission id=1, user_id=1, trip_id=1>
<Mission id=2, user_id=2, trip_id=1>
<Mission id=3, user_id=2, trip_id=2>

所以Mission.find(1).images返回3张图像，而2和3返回一个空数组.

So Mission.find(1).images gives back the 3 images, whereas 2 and 3 give an empty array.

推荐答案

我认为您可以尝试

class Mission < ActiveRecord::Base
    belongs_to :user
    belongs_to :trip
    has_many :images, ->(obj) { where("#{Image.quoted_table_name}.trip_id = ?", obj.trip_id)}, through: :user
    # or
    has_many :images, ->(obj) { where("#{Image.quoted_table_name}.user_id = ?", obj.trip_id)}, through: :trip
end

此obj代表Mission类的对象，因此我在图像关联上添加了一个条件.

This obj represents the object of Mission class, so i have added the one more condition on images associations.

Mission.first.images.to_sql
#=> "SELECT \"images\".* FROM \"images\" INNER JOIN \"users\" ON \"images\".\"user_id\" = \"users\".\"id\" WHERE (\"images\".trip_id = 1) AND \"users\".\"id\" = ?"

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