343
votes

I was wondering how to get the current URL within a template.

Say my current URL is:

.../user/profile/

How do I return this to the template?

13
possible duplicate of Reading path in templatesMark Mikofski
All the below answers made me think I needed to do some gymnastics to get access to request in a template. In Django 1.10 I just access {{request.path}} in the template and it works. By default django.core.context_processors.request is already configured in settings.py if you used startprojectUser

13 Answers

286
votes

Django 1.9 and above:

## template
{{ request.path }}  #  -without GET parameters 
{{ request.get_full_path }}  # - with GET parameters

Old:

## settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
    'django.core.context_processors.request',
)

## views.py
from django.template import *

def home(request):
    return render_to_response('home.html', {}, context_instance=RequestContext(request))

## template
{{ request.path }}
289
votes

You can fetch the URL in your template like this:

<p>URL of this page: {{ request.get_full_path }}</p>

or by

{{ request.path }} if you don't need the extra parameters.

Some precisions and corrections should be brought to hypete's and Igancio's answers, I'll just summarize the whole idea here, for future reference.

If you need the request variable in the template, you must add the 'django.core.context_processors.request' to the TEMPLATE_CONTEXT_PROCESSORS settings, it's not by default (Django 1.4).

You must also not forget the other context processors used by your applications. So, to add the request to the other default processors, you could add this in your settings, to avoid hard-coding the default processor list (that may very well change in later versions):

from django.conf.global_settings import TEMPLATE_CONTEXT_PROCESSORS as TCP

TEMPLATE_CONTEXT_PROCESSORS = TCP + (
    'django.core.context_processors.request',
)

Then, provided you send the request contents in your response, for example as this:

from django.shortcuts import render_to_response
from django.template import RequestContext

def index(request):
    return render_to_response(
        'user/profile.html',
        { 'title': 'User profile' },
        context_instance=RequestContext(request)
    )
19
votes

The below code helps me:

 {{ request.build_absolute_uri }}
7
votes

Both {{ request.path }} and {{ request.get_full_path }} return the current URL but not absolute URL, for example:

your_website.com/wallpapers/new_wallpaper

Both will return /new_wallpaper/ (notice the leading and trailing slashes)

So you'll have to do something like

{% if request.path == '/new_wallpaper/' %}
    <button>show this button only if url is new_wallpaper</button>
{% endif %}

However, you can get the absolute URL using (thanks to the answer above)

{{ request.build_absolute_uri }}

NOTE: you don't have to include request in settings.py, it's already there.

6
votes

In django template
Simply get current url from {{request.path}}
For getting full url with parameters {{request.get_full_path}}

Note: You must add request in django TEMPLATE_CONTEXT_PROCESSORS

6
votes

I suppose send to template full request is little bit redundant. I do it this way

from django.shortcuts import render

def home(request):
    app_url = request.path
    return render(request, 'home.html', {'app_url': app_url})

##template
{{ app_url }}
4
votes

The other answers were incorrect, at least in my case. request.path does not provide the full url, only the relative url, e.g. /paper/53. I did not find any proper solution, so I ended up hardcoding the constant part of the url in the View before concatenating it with request.path.

1
votes

This is an old question but it can be summed up as easily as this if you're using django-registration.

In your Log In and Log Out link (lets say in your page header) add the next parameter to the link which will go to login or logout. Your link should look like this.

<li><a href="http://www.noobmovies.com/accounts/login/?next={{ request.path | urlencode }}">Log In</a></li>

<li><a href="http://www.noobmovies.com/accounts/logout/?next={{ request.path | urlencode }}">Log Out</a></li>

That's simply it, nothing else needs to be done, upon logout they will immediately be redirected to the page they are at, for log in, they will fill out the form and it will then redirect to the page that they were on. Even if they incorrectly try to log in it still works.

1
votes

You can get the url without parameters by using {{request.path}} You can get the url with parameters by using {{request.get_full_path}}

0
votes

Above answers are correct and they give great and short answer.

I was also looking for getting the current page's url in Django template as my intention was to activate HOME page, MEMBERS page, CONTACT page, ALL POSTS page when they are requested.

I am pasting the part of the HTML code snippet that you can see below to understand the use of request.path. You can see it in my live website at http://pmtboyshostelraipur.pythonanywhere.com/

<div id="navbar" class="navbar-collapse collapse">
  <ul class="nav navbar-nav">
        <!--HOME-->
        {% if "/" == request.path %}
      <li class="active text-center">
          <a href="/" data-toggle="tooltip" title="Home" data-placement="bottom">
            <i class="fa fa-home" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true">
            </i>
          </a>
      </li>
      {% else %}
      <li class="text-center">
          <a href="/" data-toggle="tooltip" title="Home" data-placement="bottom">
            <i class="fa fa-home" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true">
            </i>
          </a>
      </li>
      {% endif %}

      <!--MEMBERS-->
      {% if "/members/" == request.path %}
      <li class="active text-center">
        <a href="/members/" data-toggle="tooltip" title="Members"  data-placement="bottom">
          <i class="fa fa-users" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true"></i>
        </a>
      </li>
      {% else %}
      <li class="text-center">
        <a href="/members/" data-toggle="tooltip" title="Members"  data-placement="bottom">
          <i class="fa fa-users" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true"></i>
        </a>
      </li>
      {% endif %}

      <!--CONTACT-->
      {% if "/contact/" == request.path %}
      <li class="active text-center">
        <a class="nav-link" href="/contact/"  data-toggle="tooltip" title="Contact"  data-placement="bottom">
            <i class="fa fa-volume-control-phone" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true"></i>
          </a>
      </li>
      {% else %}
      <li class="text-center">
        <a class="nav-link" href="/contact/"  data-toggle="tooltip" title="Contact"  data-placement="bottom">
            <i class="fa fa-volume-control-phone" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true"></i>
          </a>
      </li>
      {% endif %}

      <!--ALL POSTS-->
      {% if "/posts/" == request.path %}
      <li class="text-center">
        <a class="nav-link" href="/posts/"  data-toggle="tooltip" title="All posts"  data-placement="bottom">
            <i class="fa fa-folder-open" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true"></i>
          </a>
      </li>
      {% else %}
      <li class="text-center">
        <a class="nav-link" href="/posts/"  data-toggle="tooltip" title="All posts"  data-placement="bottom">
            <i class="fa fa-folder-open" style="font-size:25px; padding-left: 5px; padding-right: 5px" aria-hidden="true"></i>
          </a>
      </li>
      {% endif %}
</ul>

0
votes

In Django 3, you want to use url template tag:

{% url 'name-of-your-user-profile-url' possible_context_variable_parameter %}

For an example, see the documentation

0
votes

For Django > 3 I do not change settings or anything. I add the below code in the template file.

{{ request.path }}  #  -without GET parameters 
{{ request.get_full_path }}  # - with GET parameters

and in view.py pass request variable to the template file.

view.py:

def view_node_taxon(request, cid):
    showone = get_object_or_404(models.taxon, id = cid)
    context = {'showone':showone,'request':request}
    mytemplate  = loader.get_template('taxon/node.html')
    html = mytemplate.render(context)
    return HttpResponse(html)
0
votes

Use example:

 <!-- BRAND -->
        <div class="brand">
            <div class="logo">
                {% if request.get_full_path == '/' %}
                    <a href="{% url 'front:homepage' %}" class="first-logo big-logo">
                        <img src="{% static 'assets/images/logo-big.svg' %}"
                             alt="pozitiv">
                    </a>

                    <a href="{% url 'front:homepage' %}" class="second-logo mobile-logo">
                        <img src="{% static 'assets/images/logo.svg' %}"
                             alt="<?php echo WEBSITE_NAME; ?>" >
                    </a>

                {% else %}

                    <a href="{% url 'front:homepage' %}">
                        <img src="{% static 'assets/images/logo.svg' %}"
                             alt="<?php echo WEBSITE_NAME; ?>" style="width: 320px; margin: -20px 0 0 0;">
                    </a>
                {% endif %}
            </div>
        </div>