ruby-on-rails - Rails STI和强大的参数

Question

我有两种类型的客户资源:人员和公司。

routes.rb:

resources :clients
resources :people, :controller => "clients", :type => "Person"
resources :companies, :controller => "clients", :type => "Company"

clients_controller:

def new
  @client = Client.new()
  @client.type = params[:type]
end

def create
  @client = current_partner.clients.new(client_params)
  if @client.save
    redirect_to clients_path
  ...
end

...
private
def client_params
  params.require(:client).permit(:type, :partner_id, :name, :email, :phone, :cui,   :registration_id, :address)
end

def find_client
  @client ||= Client.find(params[:id])
end

client.rb

class Client < ActiveRecord::Base
  validates_presence_of :type
  CLIENT_TYPES = ['Person', 'Company']
end

人.rb

class Person < Client
  validates_presence_of :name, :email, :phone
end

compay.rb

class Company < Client
  validates_presence_of :name, :email, :cui, :registration_id, :phone
  validates_uniqueness_of :cui, :registration_id, uniqueness: {scope: :partner_id}
end

问题是当我尝试编辑客户的详细信息并提交更改时，我得到了
参数缺失或值为空:客户端。
我收到此错误的路线是.../companies/3。

对这个noobie问题有帮助吗？谢谢!

Answer 1

您可以在执行许可操作之前将继承的模型伪装成其父模型

def client_params

  if params.has_key? :person
    params[:client] = params.delete :person
  elsif params.has_key? :company
    params[:client] = params.delete :company
  end

  params.require(:client).permit(...)

end

它只是在params哈希中重命名模型的键，就可以了。