Ruby on Rails传递对yield的引用(模板) [英] Ruby on Rails pass reference to yield (template)

问题描述

简而言之:我的每个标签都有自己的形式，所以我决定制作一个单一的布局，只是将表单本身作为布局的可变内容.

To make long story short: each of my tabs has its own form, so I decided to make a single layout and just to have a forms themselves as a variable content for a layout.

但是我需要在布局中使用 form_for ，而不是在每个表单中都使用它，因为在布局中我还有一些其他常见的表单元素.

But I need to have form_for to be in a layout, rather then having it in each of the forms, because I have some other common form elements in the layout.

那么如何将表单生成器的引用 f 传递给模板?

So how can I pass the form builder's reference f to the template ?

布局代码:

<% content_for(:content) do %>
  <%= form_for current_form do |f| %>
    <%= yield %>
    <%= f.submit "Submit" %>
  <% end %>
<% end %>

有可能吗?

PS发现了这一点:干一个助手:包装form_for并访问本地表单变量(@rubish的答案)，但是<%= yield f %>似乎不起作用，对于该视图f仍未定义.

P.S Found this one: DRYing up a helper: wrap form_for and access local form variable (@rubish's answer), but <%= yield f %> doesn't seem to be working, f still remains undefined for the view.

推荐答案

为什么不为标签创建通用模板( not 布局)，并使用 partial 每个标签内容的模板?

Why don't you make a common template (not layout) for the tabs, and use a partial template for the content of each tab?

然后您可以执行以下操作:

Then you can do something like:

<%= render :partial => @tab_name, :locals => {:form => f} %>

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