Rails STI和强大的参数

Question

我有一个客户端资源有两种类型:人员和公司.

routes.rb中:

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

person.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,当我尝试编辑客户端的详细信息并提交更改时,我发现param丢失或值为空:客户端.我收到此错误的路线是....../companies/3.

对这个noobie问题的任何帮助？谢谢!

Answer 1

您可以在允许操作之前将继承的模型伪装成它的父级

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散列中重命名模型的键,并且可以解决问题.

Answer 2

楷模

我认为您没有正确使用STI

STI 是针对Models 的，而不是 针对controllers. 根据MVC 编程模式，您Models处理所有数据构造方法。您的控制器用作您的user输入和您的views:

这意味着如果你想使用或创建 STI 驱动的功能，你最好只使用后端classes（而不是手动传递type等）：

#app/models/person.rb
Class Person < Client
   ...
end

#app/models/company.rb
Class Company < Client
end

#app/models/client.rb
Class Client < ActiveRecord::Base
end

这将使您能够做到这一点：

#config/routes.rb
resources :clients
resources :people,    controller: "clients", type: "Person"
resources :companies, controller: "clients", type: "Company"

#app/controllers/clients_controller.rb
Class ClientsController < ApplicationController
    def create
       model = get_model(params[:type])

       @model = model.new(model_params)
       @model.save
    end

    private

    def get_model type
       return type.singularize.titleize.camelize.constantize
    end

    def model_params
        params.require(params[:type].to_sym).permit(:client, :model, :attributes)
    end
end

这将Model是您通过访问/people或设置的各种 STI ，/companies然后让您能够将数据保存到其中（当然，这将保存到Client模型type中）