Question

我有这个应用，其中模型User可以有多个“频道”。应用程序的渠道部分必须能够轻松扩展，每个部分都有自己的模型。

我开始创建一个Channel模型，其belongs_to User关系与User has_many Channel相反。

我在考虑使用与此类似的API：

user = User.create(name: 'John')

user.channels.create(name: 'google', account_id: '1234') # should create a GoogleChannel::Google Model
user.channels.create(name: 'facebook', account_name: 'qwerty') # should create a FacebookChannel::Facebook Model

我正在玩Channel模型，每个频道模型都有一个专用模型，用于与该模型has_one关系的每个频道（google，facebook等）。< / p>

更新

我正在使用 mongoid 和rails

Answer 1

我不确定这是否有效。它使用STI。

第一种方法：单表继承

class User << ApplicationRecord
  has_many :channels
  delegate :google_channels, :facebook_channels, :twitter_channels, to: :channels
end

class Channel << ApplicationRecord
  belongs_to :user
  self.inheritance_column = :brand

  scope :google_channels,   -> { where(brand: 'Google')   } 
  scope :facebook_channels, -> { where(brand: 'Facebook') } 
  scope :twitter_channels,  -> { where(brand: 'Twitter') }

  def self.brands
    %w(Google Facebook Twitter)
  end 
end

class GoogleChannel << Channel; end
class FacebookChannel << Channel; end
class TwitterChannel << Channel; end

我认为你可以：

current_user.channels << GoogleChannel.new(name: "First Google Channel") 
current_user.channels << Facebook.new(name: "Facebook") 
current_user.channels << Twitter.new(name: "Tweety")
current_user.channels << GoogleChannel.new(name: "Second Google Channel") 

googs = current_user.google_channels
all = current_user.channels
# etc.

所有频道共享同一张桌子。如果您需要为每个不同的品牌提供不同的属性，这将不是最佳选择。

第二种方法：多态模型

如果每个型号（品牌）需要不同的表格，您可以使用多态方法（未经测试）：

class User << ApplicationRecord
  has_many :channels
  has_many :google_channels, through: :channels, source: :branded_channel, source_type: 'GoogleChannel'
  has_many :facebook_channels, through: :channels, source: :branded_channel, source_type: 'FacebookChannel'
end

class Channel << ApplicationRecord
  belongs_to :user
  belongs_to :branded_channel, polymorphic: true
end

#This channel has its own table, and can have more attributes than Channel
class GoogleChannel << ApplicationRecord
  has_one :channel, as: :branded_channel
end

#This channel has its own table, and can have more attributes than Channel
class FacebookChannel << ApplicationRecord
  has_one :channel, as: :branded_channel
end

goog = GoogleChannel.create(all_google_channel_params)
face = GoogleChannel.create(all_facebook_channel_params)
current_user.channels << Channel.new(branded_channel: goog)
current_user.channels << Channel.new(branded_channel: face)

Answer 2

我假设你想创建一个User有很多频道的STI，所以在Rails 5你可以试试这个：

class User < ApplicationRecord
  has_many :user_channels
  has_many :channels, through: :user_channels
  has_many :facebook_channels, class_name: 'FacebookChannel', through: :user_channels
  has_many :google_channels, class_name: 'GoogleChannel', through: :user_channels
end

class UserChannel < ApplicationRecord
  # user_id
  # channel_id
  belongs_to :user
  belongs_to :channel
  belongs_to :facebook_channel, class_name: 'FacebookChannel', foreign_key: :channel_id, optional: true
  belongs_to :google_channel, class_name: 'GoogleChannel', foreign_key: :channel_id, optional: true
end

class Channel < ApplicationRecord
  # You need to have "type" column to be created for storing different chanels
  has_many :user_channels
  has_many :users, through: :user_channels
end

class FacebookChannel < Channel
  has_many :user_channels
  has_many :users, through: :user_channels
end

class GoogleChannel < Channel
  has_many :user_channels
  has_many :users, through: :user_channels
end

现在您可以执行current_user.facebook_channels.create(name: "My Facebook Channel")并使用FacebookChannel获取所有current_user.facebook_channels

基本上它与STI额外功能的常规has_many through关系一样 - 您将子模型名称存储在type模型的Channel列中。

<强>更新

对不起，我不知道我的建议不适用于MongoDB。也许您只需在channel_type表格中创建channels列，即可建立简单的channel belongs_to user关系，然后执行以下操作：

current_user.channels.create(name: "My Channel Name", channel_type: "facebook")
current_user.channels.where(channel_type: "facebook")

您可以简单地current_user.channels为用户提供所有渠道，然后如果您需要，您可以根据channel_type值对每条记录执行所需操作。您甚至可以在Channel model中创建一些范围：

class Channel < ApplicationRecord
  scope :facebook_channels, -> { where(channel_type: "facebook") }
end

然后你可以做你想要的current_user.channels.facebook_channels

channel_type列可以是字符串或整数。

此外，如果您创建visit列（例如＆＃34;布尔＆＃34;），您可以执行current_user.channels.where(visit: true)或创建scope :visit_true, -> { where(visit: true) }并执行current_user.channels.visit_true之类的操作

你说什么？

与多个模型的关系[有趣的轨道关系概念]

2 个答案: