嘿伙计们,我一直在用一个特殊的嵌套表格用例(我正在使用Rails 2.3.5)撞墙。
基本上我有项目和付款模式,看起来像这样
class Project < ActiveRecord::Base
has_many :payments
end
class Payment < ActiveRecord::Base
belongs_to :project
accepts_nested_attributes_for :project
end
我也在使用这两种资源的嵌套路由:
map.resources :projects do |project|
project.resources :payments
end
我正在尝试使用嵌套表单,以允许用户在创建新付款时修改项目的某些属性。因此,如果项目有标题,那么创建新付款的视图将如下所示:
<% form_for([@project, @payment]) do |f| %>
<% f.fields_for :project do |project_form| %>
<%= project_form.label :title %>
<%= project_form.text_field :title %>
<% end %>
<%= f.text_field :credit_card_number %>
...
<% end %>
Payments控制器非常标准:
class PaymentsController < ApplicationController
before_filter :load_project
# GET /payments/new
# GET /payments/new.xml
def new
@payment = @project.payments.build
respond_to do |format|
format.html # new.html.erb
format.xml { render :xml => @payment }
end
end
# POST /payments
# POST /payments.xml
def create
@payment = @project.payments.build(params[:payment])
respond_to do |format|
if @payment.save
flash[:notice] = 'Payment was successfully created.'
format.html { redirect_to([@project, @payment]) }
format.xml { render :xml => @payment, :status => :created, :location => @payment }
else
format.html { render :action => "new" }
format.xml { render :xml => @payment.errors, :status => :unprocessable_entity }
end
end
end
private
def load_project
@project = Project.find(params[:project_id])
end
end
我所发现的是,在新的付款表格中,我最终会得到这样的结果:
<input id="payment_project_attributes_title" name="payment[project_attributes][title]" size="30" type="text" />
<input id="payment_project_attributes_id" name="payment[project_attributes][id]" type="hidden" value="56" />
(注意自动创建的#payment_project_attributes_id)
提交表单时,rails会这样接收(请记住项目#56已经存在):
"payment"=>{"project_attributes"=>{"title"=>"test title", "id"=>"56"}, "credit_card_number"=>"41111111111111111"}
这就是问题所在:当它通过控制器运行时,它不会将title属性应用于Payment的项目。
奇怪的是,如果我删除“id”=&gt;“56”,项目的标题会更新。以下是使用控制台的示例:
ruby-1.8.7-p249 > Project.find(56)
=> #<Project id: 56, title: nil, created_at: "2010-06-18 15:58:25", updated_at: "2010-06-18 16:01:37">
ruby-1.8.7-p249 > p=Project.find(56).payments.new({"project_attributes"=>{"title"=>"my new title", "id"=>"56"}})
=> #<Payment id: nil, project_id: 56, created_at: nil, updated_at: nil>
ruby-1.8.7-p249 > p.project
=> #<Project id: 56, title: nil, created_at: "2010-06-18 15:58:25", updated_at: "2010-06-18 16:01:37">
ruby-1.8.7-p249 > p=Project.find(56).payments.new({"project_attributes"=>{"title"=>"test title"}})
=> #<Payment id: nil, project_id: 56, created_at: nil, updated_at: nil>
ruby-1.8.7-p249 > p.project
=> #<Project id: nil, user_id: nil, title: "test title", created_at: nil, updated_at: nil>
(请注意,第二次payment.new,没有ID,会导致p.project.title更新)
有人有任何想法吗?
我应该注意到,我真正想做的是一层更复杂 - 我正在尝试更新Project的user_attributes(使用项目中的belongs_to:user / accepts_nested_attributes_for:user)但我排序希望只有工作才能解决这个问题。
答案 0 :(得分:1)
似乎因为你基于@project.payments的关联调用构建方法,所以它已经知道项目id是什么。也许它不喜欢您尝试更新项目ID的事实?