Question

我使用单表继承来管理不同类型的项目。

型号：

class Project < ActiveRecord::Base
end

class SiteDesign < Project
end

class TechDesign < Project
end

从projects_controller编辑动作：

def edit
   @project = Project.find(params[:id])
end

查看edit.html.erb：

<% form_for(@project, :url => {:controller => "projects",:action => "update"}) do |f| %>
    ...
    <%= submit_tag 'Update' %>
<% end %>

更新projects_controller的操作：

def update
    @project = Project.find(params[:id])
    respond_to do |format|
      if @project.update_attributes(params[:project])
        @project.type = params[:project][:type]
        @project.save
        flash[:notice] = 'Project was successfully updated.'
        format.html { redirect_to(@project) }
        format.xml  { head :ok }
      else
        format.html { render :action => "edit" }
        format.xml  { render :xml => @project.errors, :status => :unprocessable_entity }
      end
    end
  end

然后我在编辑视图上对TechDesign条目进行了一些编辑并得到错误：

NoMethodError in ProjectsController#update

You have a nil object when you didn't expect it!
You might have expected an instance of ActiveRecord::Base.
The error occurred while evaluating nil.[]

在参数中很明显，我没有项目参数名称，而是有tech_design 参数：

{"commit"=>"Update",
 "_method"=>"put",
 "authenticity_token"=>"pd9Mf7VBw+dv9MGWphe6BYwGDRJHEJ1x0RrG9hzirs8=",
 "id"=>"15",
 "tech_design"=>{"name"=>"ech",
 "concept"=>"efds",
 "type"=>"TechDesign",
 "client_id"=>"41",
 "description"=>"tech"}}

如何解决？

Answer 1

这是您问题的根源。这是将@project设置为TechDesign对象的实例。

def edit
   @project = Project.find(params[:id])
end

您可以通过在form_for调用中指定名称项目来确保事情按您所需的方式工作。

<% form_for(:project, @project, :url => {:controller => "projects",:action => "update"}) do |f| %>
    ...
    <%= submit_tag 'Update' %>
<% end %>

Answer 2

对于Rails 3

<% form_for(@project, :as => :project, :url => {:controller => "projects",:action => "update"}) do |f| %>
...
   <%= submit_tag 'Update' %>
<% end %>

Answer 3

随机说明：如果您正在使用单表继承（STI）并忘记从子类定义中删除 initialize 方法，那么当您没有预期时，您将获得类似的“nil对象” “例外。

例如：

class Parent < ActiveRecord::Base
end

class Child < Parent
  def initialize
    # you MUST call *super* here or get rid of the initialize block
  end
end

就我而言，我使用IDE创建子类，IDE创建了initialize方法。让我永远追踪......

Answer 4

对于Rails 4，我已经确认对我来说唯一有效的方法是明确指定URL和AS参数：

<% form_for(@project, as: :project, url: {controller: :projects, action: :update}) do |f| %>
    ...
<% end %>

对我来说似乎很难看！

STI和form_for问题

4 个答案: