Rails:如何根据条件显示带或不带链接的块 (link_to_if) [英] Rails: How can I show a block with or without a link based on a condition (link_to_if)

问题描述

我有一个复杂的标签块(

、

、...)根据条件链接它.

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.

我知道 link_to_if 是这样工作的:

I know about link_to_if that works like that:

<% link_to_if condition, name, path %>

如果条件为 false，则只会呈现名称.

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

而且我知道 link_to 和 &block:

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

我想要两者的结合.接受 &block 的 link_to_if 语句，如果条件为 false，则块将在没有链接的情况下呈现.不幸的是，带有 &block 的 link_to_if 语句与 link_to 语句不同:(

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 :(

有人给我建议吗?非常感谢任何帮助

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

推荐答案

我为此编写了自己的帮助程序:

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

你可以这样使用它:

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

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