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' %}

sergzach

[Django]-Django / Comet (Push): Least of all evils?

105

Figured out another way to do it, elegant enough thanks to this answer : https://stackoverflow.com/a/17614086/34871

Given an url pattern such as:

url(r'^some-url', "myapp.myview", name='my_view_name'),

my_view_name is available to the template through request ( remember you need to use a RequestContext – which is implicit when using render_to_response )

Then menu items may look like :

<li class="{% if request.resolver_match.url_name == "my_view_name" %}active{% endif %}"><a href="{% url "my_view_name" %}">Shortcut1</a></li>
<li class="{% if request.resolver_match.url_name == "my_view_name2" %}active{% endif %}"><a href="{% url "my_view_name2" %}">Shortcut2</a></li>

etc.

This way, the url can change and it still works if url parameters vary, and you don’t need to keep a list of menu items elsewhere.

vincent

Using template tag

You can simply use the following template tag:

# path/to/templatetags/mytags.py
import re

from django import template
try:
    from django.urls import reverse, NoReverseMatch
except ImportError:
    from django.core.urlresolvers import reverse, NoReverseMatch

register = template.Library()


@register.simple_tag(takes_context=True)
def active(context, pattern_or_urlname):
    try:
        pattern = '^' + reverse(pattern_or_urlname)
    except NoReverseMatch:
        pattern = pattern_or_urlname
    path = context['request'].path
    if re.search(pattern, path):
        return 'active'
    return ''

So, in you your template:

{% load mytags %}
<nav><ul>
  <li class="nav-home {% active 'url-name' %}"><a href="#">Home</a></li>
  <li class="nav-blog {% active '^/regex/' %}"><a href="#">Blog</a></li>
</ul></nav>

Using only HTML & CSS

There is another approach, using only HTML & CSS, that you can use in any framework or static sites.

Considering you have a navigation menu like this one:

<nav><ul>
  <li class="nav-home"><a href="#">Home</a></li>
  <li class="nav-blog"><a href="#">Blog</a></li>
  <li class="nav-contact"><a href="#">Contact</a></li>
</ul></nav>

Create some base templates, one for each session of your site, as for example:

home.html
base_blog.html
base_contact.html

All these templates extending base.html with a block “section”, as for example:

...
<body id="{% block section %}section-generic{% endblock %}">
...

Then, taking the base_blog.html as example, you must have the following:

{% extends "base.html" %}
{% block section %}section-blog{% endblock %}

Now it is easy to define the actived menu item using CSS only:

#section-home .nav-home,
  #section-blog .nav-blog,
  #section-contact .nav-contact { background-color: #ccc; }

semente

You could make a context variable links with the name, URL and whether it’s an active item:

{% for name, url, active in links %}
    {% if active %}
<span class='active'>{{ name }}</span>
    {% else %}
<a href='{{ url }}'>{{ name }}</a>
    {% endif %}
{% endfor %}

If this menu is present on all pages, you could use a context processor:

def menu_links(request):
    links = []
    # write code here to construct links
    return { 'links': links }

Then, in your settings file, add that function to TEMPLATE_CONTEXT_PROCESSORS as follows: path.to.where.that.function.is.located.menu_links. This means the function menu_links will be called for every template and that means the variable links is available in each template.

Simeon Visser

I have come up with a way to utilize block tags within menu-containing parent template to achieve something like this.

base.html – the parent template:

<a href="/" class="{% block menu_home_class %}{% endblock %}">Home</a>
<a href="/about" class="{% block menu_about_class %}{% endblock %}">About</a>
<a href="/contact" class="{% block menu_contact_class %}{% endblock %}">Contact</a>

{% block content %}{% endblock %}

about.html – template for a specific page:

{% extends "base.html" %}

{% block menu_about_class %}active{% endblock %}

{% block content %}
    About page content...
{% endblock %}

As you can see, the thing that varies between different page templates is the name of the block containing active. contact.html would make use of menu_contact_class, and so on.

One benefit of this approach is that you can have multiple subpages with the same active menu item. For example, an about page might have subpages giving information about each team members of a company. It could make sense to have the About menu item stay active for each of these subpages.

Magnus Teekivi

Here ist my solution:

{% url 'module:list' as list_url %}
{% url 'module:create' as create_url %}

<ul>
    <li><a href="{% url 'module:list' %}" class="{% if request.path == list_url %}active{% endif %}">List Page</a></li>
    <li><a href="{% url 'module:create' %}" class="{% if request.path == create_url %}active{% endif %}">Creation Page</a></li>
</ul>

Tobias Ernst

Assuming the nav item is a link with the same URL as the current page, you could just use JavaScript. Here’s an annotated method that I use to add a class="active" to a li in a navigation menu with class="nav":

// Get the path name (the part directly after the URL) and append a trailing slash
// For example, 'http://www.example.com/subpage1/sub-subpage/' 
//     would become '/subpage1/'
var pathName = '/' + window.location.pathname.split('/')[1];
if ( pathName != '/' ) { pathName = pathName + '/'; }

// Form the rest of the URL, so that we now have 'http://www.example.com/subpage1/'
// This returns a top-level nav item
var url = window.location.protocol + '//' +
          window.location.host +
          pathName;
console.log(url);

// Add an 'active' class to the navigation list item that contains this url
var $links = document.querySelectorAll('.nav a');
$link = Array.prototype.filter.call( $links, function(el) {
    return el.href === url;
})[0];
$link.parentNode.className += ' active';

This method means you can simply pop it into your base template once and forget about it. No repetition, and no manual specification of the page URL in each template.

One caveat: this obviously only works if the url found matches a navigation link href. It would additionally be possible to specify a couple of special use cases in the JS, or target a different parent element as needed.

Here’s a runnable example (keep in mind, snippets run on StackSnippets):

// Get the path name (the part directly after the URL) and append a trailing slash
// For example, 'http://www.example.com/subpage1/sub-subpage/' 
//     would become '/subpage1/'
var pathName = '/' + window.location.pathname.split('/')[1];
if ( pathName != '/' ) { pathName = pathName + '/'; }

// Form the rest of the URL, so that we now have 'http://www.example.com/subpage1/'
// This returns a top-level nav item
var url = window.location.protocol + '//' +
          window.location.host +
          pathName;
console.log(url);

// Add an 'active' class to the navigation list item that contains this url
var $links = document.querySelectorAll('.nav a');
$link = Array.prototype.filter.call( $links, function(el) {
    return el.href === url;
})[0];
$link.parentNode.className += ' active';

li {
  display: inline-block;
  margin: 0 10px;
}
a {
  color: black;
  text-decoration: none;
}
.active a {
  color: red;
}

<ul class="nav">
  <li>
    <a href="http://example.com/">Example Link</a>
  </li>
  <li>
    <a href="http://stacksnippets.net/js/">This Snippet</a>
  </li>
  <li>
    <a href="https://google.com/">Google</a>
  </li>
  <li>
    <a href="http://stackoverflow.com/">StackOverflow</a>
  </li>
</ul>

rnevius

I ran into this challenge today with how to dynamically activate a “category” in a sidebar. The categories have slugs which are from the DB.

I solved it by checking to see category slug was in the current path. The slugs are unique (standard practice) so I think this should work without any conflicts.

{% if category.slug in request.path %}active{% endif %}

Full example code of the loop to get the categories and activate the current one.

{% for category in categories %}
<a class="list-group-item {% if category.slug in request.path %}active{% endif %}" href="{% url 'help:category_index' category.slug %}">
  <span class="badge">{{ category.article_set.count }}</span>
  {{ category.title }}
</a>
{% endfor %}

Alex Phelps

Based on @vincent’s answer, there is an easier way to do this without messing up with django url patterns.

The current request path can be checked against the rendered menu item path, and if they match then this is the active item.

In the following example I use django-mptt to render the menu but one can replace node.path with each menu item path.

<li class="{% if node.path == request.path %}active{% endif %}">
    <a href="node.path">node.title</a>
</li>

Wtower

I am using an easier and pure CSS solution. It has its limitations, of which I know and can live with, but it avoids clumsy CSS class selectors, like this:

<a href="index.html" class="item{% if url == request.path %}active{% endif %}">index</a>

Because a space character before active is missing the class selector gets called itemactive instead of item active and this isn’t exactly too difficult to get wrong like that.

For me this pure CSS solution works much better:

a.item /* all menu items are of this class */
{
    color: black;
    text-decoration: none;
}

a.item[href~="{{ request.path }}"] /* just the one which is selected matches */
{
    color: red;
    text-decoration: underline;
}

Notice: This even works if the URL has additional path components, because then href also matches partially. That could eventually cause ‘collisions’ with more than one match, but often enough it just works, because on well structured websites a “subdirectory” of an URL usually is a child of the selected menu item.

MaxC

I personally find the simplest way is to create blocks for each link like so:

    # base.py
...
<a href="{% url 'home:index' %}" class={% block tab1_active %}{% endblock %}>
...
<a href="{% url 'home:form' %}" class={% block tab2_active %}{% endblock %}>
...

And then in each relative template declare that link as "active" e.g.:

tab1 template:

{% block tab1_active %}"active"{% endblock %}

tab2 template:

{% block tab2_active %}"active"{% endblock %}

Josh

Simply use template tags

# app/templatetags/cores.py

from django import template
from django.shortcuts import reverse

register = template.Library()

@register.simple_tag
def active(request, url, classname):
    if request.path == reverse(url):
        return classname
    return ""

Do like this in your template

{% load cores %}
<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Example</title>
</head>
<body>
    <div>
        <a href="{% url 'myUrl' %}" class="{% active request 'myUrl' 'activeClass' %}">myUrl</a>
    </div>
</body>
</html>

Have fun

Al Mahdi

Source:stackexchange.com

[Django]-How to render menu with one active item with DRY?

Using template tag

Using only HTML & CSS

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