How to get URL of current page, including parameters, in a template?

Solution 1:

Write a custom context processor. e.g.

def get_current_path(request):
    return {
       'current_path': request.get_full_path()
     }

add a path to that function in your TEMPLATE_CONTEXT_PROCESSORS settings variable, and use it in your template like so:

{{ current_path }}

If you want to have the full request object in every request, you can use the built-in django.core.context_processors.request context processor, and then use {{ request.get_full_path }} in your template.

See:

• Custom Context Processors
• HTTPRequest's get_full_path() method.

Solution 2:

Use Django's build in context processor to get the request in template context. In settings add request processor to TEMPLATE_CONTEXT_PROCESSORS

TEMPLATE_CONTEXT_PROCESSORS = (

    # Put your context processors here

    'django.core.context_processors.request',
)

And in template use:

{{ request.get_full_path }}

This way you do not need to write any new code by yourself.

Solution 3:

In a file context_processors.py (or the like):

def myurl( request ):
  return { 'myurlx': request.get_full_path() }

In settings.py:

TEMPLATE_CONTEXT_PROCESSORS = (
  ...
  wherever_it_is.context_processors.myurl,
  ...

In your template.html:

myurl={{myurlx}}