如何在Rails中的嵌套表单中使用link_to?

时间:2013-03-26 23:03:40

标签: ruby-on-rails paperclip nested-forms nested-attributes

我有一个可能包含许多文件的对象的嵌套表单。我在创建对象时没有问题,但是当我需要编辑它时,我发现问题,表单应该有文件URL的链接,但我不知道如何引用文件本身。我正在使用回形针。在这种情况下如果只有一个文件并且它在对象模型中我会做类似@ object.file.url的事情,但由于有很多文件,我不知道如何继续这种情况......

edit.html.haml

= form_for @object, :url => { :action => "update", :controller => "object", :id => @object.id, :method => "post" }, :html => { :multipart => true }  do |f|

  - render "object/object_form", object: f.object, f: f

object_form.html.haml

.field
  = f.label :name
  = f.text_field :name

= f.fields_for :files do |builder|
  = render 'object/file_fields', :f => builder
%p= link_to_add_fields "Add new file", f, :file

.field
  = f.submit "Save"

file_fields.html.haml

%fieldset
  .field
    = f.label :file
    = f.file_field :file
  .field
    = f.hidden_field :_destroy
    = link_to "Remove File", '#', class: "remove_fields"

我正在尝试做的事情显然不适用于file_fields.html.haml

%fieldset
  .field
    = f.label :file
    = f.file_field :file
    = link_to f.file.url 
  .field
    = f.hidden_field :_destroy
    = link_to "Remove File", '#', class: "remove_fields"

1 个答案:

答案 0 :(得分:1)

我认为你应该可以做到

f.object.url

所以:

%fieldset
  .field
    = f.label :file
    = f.file_field :file
    # Not sure if it would produce a Nil error on the new form.
    # given how the f object wouldn't yet have a url
    - if f.object.url
       = link_to f.object.url 
  .field
    = f.hidden_field :_destroy
    = link_to "Remove File", '#', class: "remove_fields"