突破铁轨中的条件障碍

Question

I have to break out of the if condition if the index is greater than 5 after appending the icon in my row. Right now it keeps on adding the icon.

<div class="row sub-navigation">
  <% @projects.each_with_index do |project, index| %>       
  <div class="col-sm-2 col-xs-1">                                  
    <% if index > 5 %>
      <%= link_to "", path, remote: true, id: "project_div", 
                class: "glyphicon glyphicon-chevron-down" %>
    <% else %>      
      <%= link_to project.name, project_url(project.project_id), class: ('active' if current_page?(project_path(project.project_id)) ) %>
    <% end %>
  </div>
  <% end %>
</div>

Answer 1

You can use break

<% @projects.each_with_index do |project, index| %>
   <% break if index < 5 %>
<% end %>

--

Or you could also use .take as per this answer :

<% @projects.take(5).each do |project| %>
   ...
<% end %>

This will allow you limit the value of the loop to 5 objects only, preventing the need for further logic

---

break is a common programming function, designed to get out of a loop

What Dax and I were suggesting was to add it along-side your if statement:

<% if index < 5 %>
    <% break %>
<% else %>
     ... do something
<% end %>

If you just want to add an icon for the first 5 links, you'll want to do this:

<% your_class = index > 5 ? nil : "icon_class" %>
<%= link_to "path", path_helper, class: your_class %>

--

Update

In response to your pastie, here's what you need to do:

<div class="row sub-navigation">
  <% @projects.each_with_index do |project, index| %>       
  <div class="col-sm-2 col-xs-1">               
    <% your_class = index > 5? "icon_class" : nil %>
    <% link_to "", path, remote: true, id: "project_div", class: your_class %>
  </div>
</div>

Answer 2

您可以尝试<% break if index > 5 %>