我的页面上有一个过滤器。
<ul>
<li>
<%= link_to "hot", root_path(:filter => ["hot"]) %>
</li>
<li>
<%= link_to "recent", root_path(:filter => ["recent"]) %>
</li>
<li>
<%= link_to "most popular", root_path(:filter => ["popular","day"]) %>
</li>
<li>
<%= link_to "most liked", root_path(:filter => ["liked","day"]) %>
</li>
</ul>
另一个页面上的几乎完全相同的过滤器(除了路径)(对于后续用户):
<ul>
<li>
<%= link_to "hot", followed_index_path(:filter => ["hot"]) %>
</li>
<li>
<%= link_to "recent", followed_index_path(:filter => ["recent"]) %>
</li>
<li>
<%= link_to "most popular", followed_index_path(:filter => ["popular","day"]) %>
</li>
<li>
<%= link_to "most liked", followed_index_path(:filter => ["liked","day"]) %>
</li>
</ul>
什么是将这些分成一部分的最佳方法?现在我对每个版本都有一个部分,这对我来说似乎很麻烦。
编辑:
我按照以下方式工作:
<ul>
<li>
<%= link_to "hot", url_for(:action => "index", :filter => ["hot"]) %>
</li>
<li>
<%= link_to "recent", url_for(:action => "index", :filter => ["recent"]) %>
</li>
<li>
<%= link_to "most popular", url_for(:action => "index", :filter => ["popular","day"]) %>
</li>
<li>
<%= link_to "most liked", url_for(:action => "index", :filter => ["liked","day"]) %>
</li>
</ul>
答案 0 :(得分:-1)
您可以尝试实施帮助方法。