How to render menu with one active item with DRY?

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.

score 104 · Answer 1 · edited May 23 '17 at 12:25

104

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.

edited May 23 '17 at 12:25

Community

1
1

answered Dec 09 '14 at 12:35

vincent

6,368
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1

just wanted to say that this is perfect because it allows you to use this in a single 'global' template instead of implementing the same thing several times in individual template files. thanks – Jack Sep 02 '15 at 22:20
1

`request.resolver_match.url_name` does not return a namespaced name. – Flimm May 13 '16 at 12:09
6

use `request.resolver_match.view_name` for namespaced name – zhenming Jul 13 '16 at 04:55
6

This is an excellent solution. Much easier and more elegant than the others. – larapsodia Aug 29 '16 at 01:43
This won't work if you have submenus and want the parent '
' item to have the 'active' class. (It would work if you expand the if statement to accept all sub-views but you don't want to go down that route.)

kristian

Sep 17 '16 at 20:13

2

This answer is the cleanest to implement. I read many posts on the forum and they are pulling off complicated solutions, but for most of us that are not going to develop a very big app this works just like a charm. I'd like to add one thing: most likely, one will need to have 2 class names in the html, one for styling the menu items if they are not selected, and one for the case they are selected/active. I give an example in the next comment. – Ibo Sep 29 '17 at 17:44

1

Example applying 2 classes to meny items: `{% else %}` template tag would suffice: `

Home

` – Ibo Sep 29 '17 at 17:47

score 58 · Answer 2 · edited Jul 10 '19 at 11:04

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; }

I wanted exact match, so I modified the pattern to: `pattern = '^' + reverse(pattern_or_urlname) + '$'`. — daigorocub, Aug 07 '14 at 10:49
@daigorocub so you broke the example below. You can provide a ^ like in `{% active '^/regex/' %}`. I think we must revert your edit. ;) — semente, Aug 08 '14 at 13:40
Well, this post really helped me but I had to add the $ sign for my code to work and I believe it adds something to the answer, no?. I didn't use the whole example - just needed the first part. Thanks. — daigorocub, Aug 09 '14 at 10:44
Thank you for inspiration. I have modified your code to work even with I18N pattern, that has language prefixed path. You can edit your comment if you like, so that others can benefit from this. http://pastebin.com/8hegST6b — darkless, Aug 20 '14 at 19:12
The template tag method is nice! I added an `exact` parameter to the `active` tag, adding a trailing `$` to the regex, because when you're linking to the index url '/', the link would always be active, since all urls start with `/` — phoibos, Jun 18 '17 at 12:55

score 34 · Accepted Answer · edited May 31 '17 at 23:27

34

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' %}

edited May 31 '17 at 23:27

dan-klasson

13,734
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answered Mar 21 '12 at 09:08

sergzach

6,578
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How to incorporate this approach if I use `{% extends "includes/tabs.html" %}` to create desired layouts? – Evgeny Bobkin Jan 03 '18 at 11:41
1

Why we should include tabs in every page template? It is not DRY solution, I think... – Abdüssamet Aslan Aug 12 '18 at 13:03
1

It makes totally sense. From the including template you can even choose different menu templates on per-page basis, for example. – jalone Apr 02 '20 at 23:43
@EvgenyBobkin Do you want to specify a menu (by using `extend`)? If you use `extend` then probably a menu is already in a base template and it should not be called 'tabs.html'. In your base template you can handle `active_tab` by assigning to a variable. `extend` is to specify a parent template. – sergzach Aug 13 '20 at 16:41

score 4 · Answer 4 · answered Jan 25 '18 at 11:15

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.

Simeon Visser · Answer 5 · 2012-03-20T19:36:49.470

4

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.

edited Mar 20 '12 at 19:36

answered Mar 20 '12 at 19:31

Simeon Visser

118,920
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What do you mean for 3 variables in *for* loop? for **name**, **url**, **active**. Is this correct to use 3 variables? – sergzach Mar 20 '12 at 19:45
The `name` is the name of the link (for example, Home), the `url` is the URL to where the link goes and `active` is whether the page is currently active. You can compute this for each page and you would get a list of links, such as `[("Home", "/home", True), ("Community", "/community", False), ("About", "/about", False)]`. – Simeon Visser Mar 20 '12 at 19:48
1

It´s indeed possible to use three variables. You can read the documentation of a for loop in Django here: https://docs.djangoproject.com/en/1.3/ref/templates/builtins/#for (see their example with points). – Simeon Visser Mar 20 '12 at 20:02
This is exactly the solution I came up with. It seems the elegance of python and the power of context processors often goes out the window when it comes to templates, so I like this solution. – andyhasit Jul 02 '17 at 00:37

score 3 · Answer 6 · answered Jan 03 '19 at 00:02

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>

rnevius · Answer 7 · 2015-11-24T15:09:07.093

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>

score 1 · Answer 8 · answered Feb 06 '16 at 03:29

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 %}

I can see this go wrong. For instance, one slug being `amazing`, and another being `amazing_stuff`. The unique constraint is satisfied, and yet `category.slug in request.path` will be true on both pages for `amazing`. — Flimm, May 13 '16 at 12:11

score 0 · Answer 9 · answered Jul 16 '15 at 11:50

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>

score 0 · Answer 10 · answered Oct 21 '16 at 01:20

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.

score 0 · Answer 11 · answered Aug 13 '20 at 10:39

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 %}

score 0 · Answer 12 · answered Sep 07 '21 at 07:51

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

How to render menu with one active item with DRY?

12 Answers12

Using template tag

Using only HTML & CSS

Linked