Question

我正在尝试在创建新条目时干掉我的某个表单，这些表单有时会有其他字段（对于嵌套模型），但在执行更新时不需要这些字段。因此，我试图通过一个块传递其他字段，但表单对象没有正确传递。

有没有办法将表单对象（或任何变量）传递给yield？

参考示例代码：

_form.slim

= nested_form_for @model do |f|
  .row
    = f.label :name
    = f.text_field :name, autofocus: true
  ...
  = yield
  ...
  = f.submit 'Save'

new.html.slim

== render layout: 'model/form' do
  h3 Additional Fields
  = f.fields_for :nested do |h|
      = a.label :name, 'Nested Name'
      = a.text_field :name
      = a.link_to_remove do
        = fa_icon 'times-circle-o'
  = f.link_to_add "Add another nested model", :nested

edit.html.slim

== render layout: 'model/form'

Answer 1

详细说明我的评论，这就是我使用partials的方式：

<强> _form.slim

= nested_form_for @model do |f|
  .row
    = f.label :name
    = f.text_field :name, autofocus: true

  ...

  - if defined?(additional_fields)
    h3 Additional Fields
      = f.fields_for :nested do |h|
          = a.label :name, 'Nested Name'
          = a.text_field :name
          = a.link_to_remove do
            = fa_icon 'times-circle-o'
      = f.link_to_add "Add another nested model", :nested

  ...

  = f.submit 'Save'

<强> new.html.slim

  == render 'model/form', :additional_fields => true

<强> edit.html.slim

  == render 'model/form'

我可能会遗漏一些东西，但我不确定为什么这不起作用。

Rails 4产生额外的表单字段

1 个答案: