Question

我试图将继承与经典＆＃34;类型＆＃34;属性。在初始化用户类型：＆＃34; Admin＆＃34;我希望有一个管理对象，但它会返回用户对象。我怎样才能使一个类像一个模型一样行动，并使它像STI一样行动？

class User
  include ActiveModel::Model
  attr_accessor :type

  def persisted?
    false
  end
end

class Admin < User
end

user = User.new(type: "Admin") #=> #<User:0x007ff68a1ade60 @type="Admin">

Answer 1

class User
  include ActiveModel::Model

  def self.build(type: 'User')
    klass = Kernel.const_get(type)
    if klass.ancestors.include?(User)
      klass.new
    else
      raise "you are just not my type"
    end
  end

  def persisted?
    false
  end
end

class Admin < User
end

> u = User.build(type: 'Admin) # => instance of Admin
> u = User.build # => instance of User
> u = User.build(type: 'Object') # => RuntimeError

ActiveModel :: Model继承不按预期工作

1 个答案: