如何为具有相同关系的两个外键的模型编写关联？

Question

我又回到了一个我自己没有解决的问题。我有一个

shipments

表，其中有两列

addresses

，即

to_address_id

和

from_address_id

。

在查看了相关问题（Same Model for Two Been_to Associations 迁移）后，我仍然很困惑。这是我到目前为止的工作。我还没有运行迁移。我试图在头脑中解决这个问题，然后再为此而烦恼并浪费一两个小时。我真的很感谢你的帮助。

class AddAddressFromAndAddressToToAddress < ActiveRecord::Migration[6.0]
  def change
    add_reference :address, :address_from, null: false
    add_reference :address, :address_to, null: false
    add_foreign_key :address, :shipments, column: :to_address_id
    add_foreign_key :address, :shipments, column: :from_address_id
  end
en

Answer 1

如果我理解正确的话，您想在发货中添加始发地和目的地地址，对吗？

在这种情况下：

如果您想创建迁移来创建关系

class AddOriginAndDestinyAddressToShipment < ActiveRecord::Migration[6.0]
  def change
    add_reference :shipments, :from_address, foreign_key: { to_table: :addresses }
    add_reference :shipments, :to_address, foreign_key: { to_table: :addresses }
  end
end

如果你想创建一个有关系的模型

class CreateShipments < ActiveRecord::Migration[6.0]
  def change
    create_table :shipments do |t|
      t.datetime :deadline
      t.references :from_address, null: false, foreign_key: { to_table: :addresses }
      t.references :to_address, null: false, foreign_key: { to_table: :addresses }
      t.timestamps
    end
  end
end

模型发货必须是这样的

class Shipment < ApplicationRecord
  belongs_to :to_address, class_name: 'Address'
  belongs_to :from_address, class_name: 'Address'
end

Answer 2

我不需要迁移，因为外键已经就位。我尝试的那个不起作用，

strong_migrations

宝石正在抱怨。我只需要将其正确连接即可。

我想展示如何避免一些混乱，因此我可以输入

shipment.to_address

或

shipment.address_to

并获得相同的结果。谢谢你。

class Shipment < ApplicationRecord
  belongs_to :address_from, class_name: 'Address', foreign_key: :from_address_id
  belongs_to :from_address, class_name: 'Address', foreign_key: :from_address_id
  belongs_to :address_to,   class_name: 'Address', foreign_key: :to_address_id
  belongs_to :to_address,   class_name: 'Address', foreign_key: :to_address_id
end

class Address < ApplicationRecord
  has_many :shipments
end