CODEWITHSUNDEEP
pythondjangodjango-templates
hln
hln

Reputation: 1106

Dynamic active menu items Django

I have created a template tag for return active or nothing to every menu item, if I hardcode every url , it will work, like here:

 <li class="{% active request '^/danmark/$' %}"><a href="/danmark/"> Hele Danmark</a>

But if i use dymanic url, there wont be add any active class to any menu item, if the the url is ^/danmark/{{a.area}}/$, like here:

 {% for a in all_areas %}
        <li class="{% active request '^/danmark/{{a.area}}/$' %} "><a href= "/danmark/{{a.area}}/">{{a.area}} </a></li>     
    {% endfor %}

and template tag:

from django import template register = template.Library()

@register.simple_tag
def active(request, pattern):
   import re
   if re.search(pattern, request.path):
      return 'active'
   return ''

What can I do here?

Upvotes: 1

Views: 3087

Answers (2)

Shariful Islam
Shariful Islam

Reputation: 199

You can try this 

{% with request.resolver_match.url_name as url_name %}
<ul id="menu">
    <li class="{% if url_name == 'home' %}active{% endif %}">Home</li>
    <li class="{% if url_name == 'about' %}active{% endif %}">About</li>
    <li class="{% if url_name == 'employer' %}active{% endif %}">Employer</li>
</ul>
 {% endwith %}

Where your urls are like 

url(r'$','home_view', name='home'),
url(r'about/$','about_view', name='about'),
url(r'employer/$','employer_view', name='employer')

Upvotes: 6

Goran
Goran

Reputation: 6824

Why don't put some signal in your view like:

active = 'danmark'

then in temlate you can do:

<li{% if active=='danmark' %} class="active"{% endif %}><a href="/danmark/"> Hele Danmark</a></li>

Upvotes: 0

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