Question

I would like to render a constructions like:

<a href='/home'>Home</a>
<span class='active'>Community</span>
<a href='/about'>About</a>

Where Community is selected menu item. I have menu with same options for several templates but I would not like to create combinations for each template:

<!-- for Home template-->
        <span class='active'>Home</span>
        <a href='/comminuty'>Community</a>
        <a href='/about'>About</a>
    ...
<!-- for Community template-->
        <a href='/home'>Home</a>
        <span class='active'>Community</span>
        <a href='/about'>About</a>
    ...
<!-- for About template-->
        <a href='/home'>Home</a>
        <a href='/community'>Community</a>
        <span class='active'>About</span>

We have permanent list of menu items, so, it can be more effective way – to create only one generalized structure of menu then render menu with required option for template.

For example it could be a tag that allows to do that.

Answer 1

I found easy and elegant DRY solution.

It’s the snippet:
http://djangosnippets.org/snippets/2421/

**Placed in templates/includes/tabs.html**

<ul class="tab-menu">
    <li class="{% if active_tab == 'tab1' %} active{% endif %}"><a href="#">Tab 1</a></li>
    <li class="{% if active_tab == 'tab2' %} active{% endif %}"><a href="#">Tab 2</a></li>
    <li class="{% if active_tab == 'tab3' %} active{% endif %}"><a href="#">Tab 3</a></li>
</ul>

**Placed in your page template**

{% include "includes/tabs.html" with active_tab='tab1' %}