Rails: How can I show a block with or without a link based on a condition (link_to_if)

问题

I have a complex block of tags (<h3>, <p>, ...) that I want to render with a link or without a link around it based on a condition.

I know about link_to_if that works like that:

<% link_to_if condition, name, path %>

if the condition is false only the name will be rendered.

And I know about the link_to with &block:

<% link_to path do %>
  [complex content]
<% end %>

I want a combination of both. A link_to_if statement that accepts a &block, so that the block will be rendered without a link around it, if the condition is false. Unfortunately the link_to_if statement with a &block works not like the link_to statement :(

Does anyone have suggestion for me? Any help is highly appreciated

回答1:

I wrote my own helper for this:

   def link_to_if_with_block condition, options, html_options={}, &block
     if condition
       link_to options, html_options, &block
     else
       capture &block
     end
   end

You can use it like this:

<%= link_to_if_with_block true, new_model_path { "test" } %>
<%= link_to_if_with_block true, new_model_path do %>
  Something more complicated
<% end %>

回答2:

I just overwrote the built in method cause the block use they offer really doesn't make much sense for our use. Just add it to a helper and this will make link_to_if work just like link_to.

def link_to_if(*args,&block)
  args.insert 1, capture(&block) if block_given?

  super *args
end

来源：https://stackoverflow.com/questions/10305180/rails-how-can-i-show-a-block-with-or-without-a-link-based-on-a-condition-link