Ruby on rails 使用两个外键的作用域

我有以下模式：

我想为两个外键（

author\u id

和

editor\u id

）以及单独的外键（例如

author\u propositions

和

editor\u propositions

）调用

propositions

），我需要有延迟或快速加载它们的选项（例如

User.includes（：propositions）

或不带它的

连接

）

更新：

#I have the scopes which is like this:
class User < ActiveRecord::Base
  has_many :author_proposals, class_name: 'Proposal', foreign_key: :author_id
  has_many :editor_proposals, class_name: 'Proposal', foreign_key: :editor_id
end

class Proposal < ActiveRecord::Base
  belongs_to :author, class_name: 'User', foreign_key: :author_id
  belongs_to :editor, class_name: 'User', foreign_key: :editor_id
end

#我有如下范围：
类用户

---

但我需要一个通用的，它会给我所有的建议（包括

作者建议

和

编辑建议

），它也会加载它们。我是否应该使用

上的条件有许多
？
如下设置您的关联：

class User < ActiveRecord::Base
  has_many :author_proposals, :class_name => "Proposal", :foreign_key => "author_id"
  has_many :editor_proposals, :class_name => "Proposal", :foreign_key => "editor_id"
end

class Proposal < ActiveRecord::Base
  belongs_to :author, :class_name => 'User', :foreign_key => "author_id"
  belongs_to :editor, :class_name => 'User', :foreign_key => "editor_id"
end

class User < ActiveRecord::Base
  has_many :authored_proposals, class_name: 'Proposal', foreign_key: :author_id
  has_many :editored_proposals, class_name: 'Proposal', foreign_key: :editor_id

  def proposals
    Proposal.where('author_id = :id OR editor_id = :id', { id: id }).distinct
  end
end

class Proposal < ActiveRecord::Base
  belongs_to :author, class_name: 'User', foreign_key: :author_id
  belongs_to :editor, class_name: 'User', foreign_key: :editor_id

  def users
    User.where(id: [author_id, editor_id].uniq)
  end
end

class用户“建议”；：外键=>“作者id”
有很多：编辑建议，：类名称=>“建议”；：外键=>“编辑id”
结束
课程提案'User'，：foreign\u key=>“author\u id”
属于\u to:editor，：class\u name=>'User'，：foreign\u key=>“editor\u id”
结束
我会这样做：

class User < ActiveRecord::Base
  has_many :author_proposals, :class_name => "Proposal", :foreign_key => "author_id"
  has_many :editor_proposals, :class_name => "Proposal", :foreign_key => "editor_id"
end

class Proposal < ActiveRecord::Base
  belongs_to :author, :class_name => 'User', :foreign_key => "author_id"
  belongs_to :editor, :class_name => 'User', :foreign_key => "editor_id"
end

class User < ActiveRecord::Base
  has_many :authored_proposals, class_name: 'Proposal', foreign_key: :author_id
  has_many :editored_proposals, class_name: 'Proposal', foreign_key: :editor_id

  def proposals
    Proposal.where('author_id = :id OR editor_id = :id', { id: id }).distinct
  end
end

class Proposal < ActiveRecord::Base
  belongs_to :author, class_name: 'User', foreign_key: :author_id
  belongs_to :editor, class_name: 'User', foreign_key: :editor_id

  def users
    User.where(id: [author_id, editor_id].uniq)
  end
end

class用户
您可以执行以下操作：

class User < ActiveRecord::Base
  has_many :authored_proposals, class_name: 'Proposal', foreign_key: :author_id
  has_many :editored_proposals, class_name: 'Proposal', foreign_key: :editor_id

  def proposals
    authored_proposals | editored_proposals
  end
end

class Proposal < ActiveRecord::Base
  belongs_to :author, class_name: 'User', foreign_key: :author_id
  belongs_to :editor, class_name: 'User', foreign_key: :editor_id

  def users
    author | editor
  end
end

发布您为此编写的代码我没有代码，这就是我要问的！：）首选的方法是什么？它是否具有范围并通过lambda？是否使用了
条件
？谢谢。这是针对简单的
作者建议
和
编辑建议
，但我需要这些建议，它将找到所有建议。这是我最初的想法，我想创建一个新方法，但我需要急切地加载它们。举个例子，这一个不起作用：User.includes（：propositions）。首先，我觉得这不是rails的做法。为什么不干脆
User.includes（：authored\u propositions，：edited\u propositions）
？好主意，我可以添加以下内容：
范围：propositions，->（User）{User.includes（：authored\u propositions，：edited\u propositions）}
但当我调用
User.proposements
时，它将始终被加载。如果没有其他rails方法来实现这一点，我会感到惊讶。