ruby-on-rails - ActiveRecord has_one 其中关联模型有两个 belongs_to 关联

Question

我有两个以这种方式相互关联的 ActiveRecord 模型：

class Address < ApplicationRecord
  has_one :user, class_name: User.name
end

class User < ApplicationRecord
  belongs_to :home_address, class_name: Address.name
  belongs_to :work_address, class_name: Address.name
end

用户 - > 地址关联工作正常：

home_address = Address.new
#=> <Address id:1>

work_address = Address.new
#=> <Address id:2>

user = User.create!(home_address: home_address, work_address: work_address)
#=> <User id:1, home_address_id: 1, work_address_id: 2>

user.home_address
#=> <Address id:1>

user.work_address
#=> <Address id:2>

我遇到的问题是让Address'shas_one正常工作。起初我得到一个错误 that User#address_id does not exist，这是有道理的，因为这不是外键字段的名称。要么是要么 （我通过迁移添加了这些 FK）。但是我不确定如何让它知道要使用哪个地址，直到我了解到您可以将范围传递给声明：home_address_idwork_address_idhas_one

class Address < ApplicationRecord
  has_one :user,
    ->(address) { where(home_address_id: address.id).or(where(work_address_id: address.id)) },
    class_name: User.name
end

但这会返回与以前相同的错误：Caused by PG::UndefinedColumn: ERROR: column users.address_id does not exist. 这令人困惑，因为在该范围内我没有声明我正在查看address_id. 我猜has_one隐含地有一个foreign_key :address_id，但我不知道如何设置它，因为从技术上讲有两个，:home_address_id 和:work_address_id。

我觉得我在这里很近 - 我该如何解决这个 has_one 关联？

更新

我的直觉说这里的解决方案是创建一个user方法来执行我要运行的查询，而不是声明一个has_one. 如果has_one支持此功能会很棒，但如果不支持，我会退回到那个。

class Address < ApplicationRecord
  def user
    User.find_by("home_address_id = ? OR work_address_id = ?", id, id)
  end
end

解决方案

感谢下面的@max！我最终根据他的回答提出了解决方案。我还使用了gem，它将在模型Enumerize中发挥作用。Address

class AddAddressTypeToAddresses < ActiveRecord::Migration[5.2]
  add_column :addresses, :address_type, :string
end

class User < ApplicationRecord
  has_many :addresses, class_name: Address.name, dependent: :destroy
  has_one :home_address, -> { Address.home.order(created_at: :desc) }, class_name: Address.name
  has_one :work_address, -> { Address.work.order(created_at: :desc) }, class_name: Address.name
end

class Address < ApplicationRecord
  extend Enumerize

  TYPE_HOME = 'home'
  TYPE_WORK = 'work'
  TYPES = [TYPE_HOME, TYPE_WORK]

  enumerize :address_type, in: TYPES, scope: :shallow
  # Shallow scope allows us to call Address.home or Address.work

  validates_uniqueness_of :address_type, scope: :user_id, if: -> { address_type == TYPE_WORK }
  # I only want work address to be unique per user - it's ok if they enter multiple home addresses, we'll just retrieve the latest one. Unique to my use case.
end

标签： ruby-on-railsactiverecord

Answer 1

Rails 中的每个关联只能有一个外键，因为您需要的是 SQL：

JOINS users 
ON users.home_address_id = addresses.id OR users.work_address_id = addresses.id

在这里使用 lambda 为关联添加默认范围将不起作用，因为 ActiveRecord 实际上并没有让您在关联级别上如何加入。如果您考虑它生成多少不同的查询以及该功能会导致的边缘情况的数量，这是完全可以理解的。

如果你真的想在你的用户表上有两个不同的外键的兔子洞，你可以用单表继承来解决它：

class AddTypeToAddresses < ActiveRecord::Migration[6.1]
  def change
    add_column :addresses, :type, :string
  end
end

class User < ApplicationRecord
  belongs_to :home_address, class_name: 'HomeAddress'
  belongs_to :work_address, class_name: 'WorkAddress'
end

class HomeAddress < Address
  has_one :user, foreign_key: :home_address_id
end

class WorkAddress < Address
  has_one :user, foreign_key: :work_address_id
end

但我会将外键放在另一张表上并使用一对多关联：

class Address < ApplicationRecord
  belongs_to :user
end

class User < ApplicationRecord
  has_many :addresses
end

这使您可以根据需要添加任意数量的地址类型，而不会破坏用户表。

如果您想将用户限制为一个家庭和一个工作地址，您可以这样做：

class AddTypeToAddresses < ActiveRecord::Migration[6.1]
  def change
    add_column :addresses, :address_type, :integer, index: true, default: 0
    add_index :addresses, [:user_id, :address_type], unique: true
  end
end

class Address < ApplicationRecord
  belongs_to :user
  enum address_type: {
    home: 0,
    work: 1
  }
  validates_uniqueness_of :type, scope: :user_id
end    

class User < ApplicationRecord
  has_many :addresses
  has_one :home_address,
    -> { home },
    class_name: 'Address'
  has_one :work_address,
    -> { work },
    class_name: 'Address'
end