How to get the current URL within a Django template?

Question

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?

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 startproject — User
– User, Commented Feb 21, 2017 at 11:48

Aleksei Khatkevich · Accepted Answer · 2019-05-02 16:48:57Z

378

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 }}

edited May 2, 2019 at 16:48

Aleksei Khatkevich

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answered Dec 8, 2011 at 13:53

httpete

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5 Comments

RedGlyph Over a year ago

A bit laconic, and not correct. It's render_to_response, and not render_to_request. And you can't define TEMPLATE_CONTEXT_PROCESSORS as you do in settings.py, without mentioning the other default processors that may well be used in the templates!

Routhinator Over a year ago

As of 2016, you no longer need to add anything to views.py. As long as django.core.context_processors.request is loaded in TEMPLATE_CONTEXT_PROCESSORS - you have access to {{ request.path }} from the template.

Flimm Over a year ago

request.path does not include query parameters like ?foo=bar. Use request.get_full_path instead.

Marshall X Over a year ago

@Routhinator agree with you. but it's nice to know that those middleware needs to be included to make this happen.

Tadeo Over a year ago

As commented in other answers by @Quentin and @user you should include urlencode if you plan to use it in a link, typically: .../accounts/login?next={{ request.get_full_path | urlencode }}...

RedGlyph · Accepted Answer · 2020-08-16 10:18:57Z

305

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)
    )

edited Aug 16, 2020 at 10:18

answered Apr 15, 2012 at 14:03

RedGlyph

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7 Comments

Bobort Over a year ago

I used an extended generic class view, and it was unnecessary to add request to the context.

user Over a year ago

Definitely cleaner to avoid hard coding the TCP list, but docs.djangoproject.com/en/dev/topics/settings/#default-settings says : Note that a settings file should not import from global_settings, because that’s redundant

Richard de Wit Over a year ago

return render(request, 'user/profile.html', {'title': 'User profile'}) is shorter

user Over a year ago

remember to include urlencode i.e. {{request.get_full_path|urlenode}} if you are redirecting

numerah Over a year ago

how to get parameters from get_full_path ??

|

Angel F Syrus · Accepted Answer · 2019-11-12 08:11:12Z

45

The below code helps me:

 {{ request.build_absolute_uri }}

edited Nov 12, 2019 at 8:11

Angel F Syrus

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answered Nov 12, 2019 at 6:37

Евгений Шабин

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2 Comments

waverider Over a year ago

This is useful, as it includes hostname/domain as well.

Desert Camel Over a year ago

and that is the required one for complete path or link

Mujeeb Ishaque · Accepted Answer · 2022-08-29 08:17:59Z

20

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 requests in settings.py, it's already there.

edited Aug 29, 2022 at 8:17

answered Nov 16, 2019 at 14:15

Mujeeb Ishaque

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Comments

Savad KP · Accepted Answer · 2016-01-15 07:25:47Z

6

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

answered Jan 15, 2016 at 7:25

Savad KP

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Comments

Kalob Taulien · Accepted Answer · 2020-02-28 03:19:32Z

6

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 }}

edited Feb 28, 2020 at 3:19

Kalob Taulien

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answered Oct 19, 2016 at 16:23

Radren

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Comments

CoderGuy123 · Accepted Answer · 2017-06-13 01:33:09Z

4

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.

answered Jun 13, 2017 at 1:33

CoderGuy123

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1 Comment

dotty Over a year ago

Look at the date. The answers were given 6 or 7 years ago.

Darwin · Accepted Answer · 2020-11-02 10:01:48Z

3

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)

answered Nov 2, 2020 at 10:01

Darwin

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Comments

Raj Kalathiya · Accepted Answer · 2020-12-21 05:33:04Z

3

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

edited Dec 21, 2020 at 5:33

answered Dec 18, 2020 at 16:34

Raj Kalathiya

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Comments

Hamed Shiani · Accepted Answer · 2021-08-21 16:18:49Z

3

if you are using render partial it's better to use

{{ request.path_info }}

It returns the exact url of the page

edited Aug 21, 2021 at 16:18

answered Aug 21, 2021 at 16:02

Hamed Shiani

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1 Comment

Conor Over a year ago

How can this be coded in python code? I need the url inside a PDF, not html

Flimm · Accepted Answer · 2015-12-31 13:45:10Z

1

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.

edited Dec 31, 2015 at 13:45

Flimm

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answered Jan 18, 2015 at 20:48

Chris Hawkes

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1 Comment

Quentin Over a year ago

you should encode the path if it is in a url: {{ request.path|urlencode }}

hygull · Accepted Answer · 2017-12-31 01:04:09Z

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>

A small suggestion -- if all you are doing is checking whether to add the active class to each li element, why not just do that inline within one li element: <li class="{% if "/contact/" == request.path %}active {% endif %}text-center">....</li> instead of a giant if/else block for the whole li? That would save a whole bunch of redundant code :)

Janne · Accepted Answer · 2020-06-16 23:57:30Z

0

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

answered Jun 16, 2020 at 23:57

Janne

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Comments

Serhii Zelenchuk · Accepted Answer · 2021-05-19 11:39:54Z

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>

Walk · Accepted Answer · 2021-12-09 11:17:13Z

0

To pass the get parameter to template, you can pass it via views.py file

Example:

return render(request, 'test.html',{'get_parameter_name' : get_parameter_value})

And use in template like:

{{get_parameter_name}}

answered Dec 9, 2021 at 11:17

Walk

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Comments

onapte · Accepted Answer · 2022-03-02 11:41:51Z

0

If you are using js in the template as well, you can do this:
In you js

document.addEventListener('DOMContentLoaded', function() {
    ...
    getUrl();
}

function getUrl() {
    let someDiv = document.querySelector(`#someDiv`);
    someDiv.innerHTML = window.location.href;
}

for your template

...
<div id="someDiv"><div>
...

answered Mar 2, 2022 at 11:41

onapte

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