Django: get URL of current page, including parameters, in a template

Question

Is there a way to get the current page URL and all its parameters in a Django template?

For example, a templatetag that would print full URL like /foo/bar?param=1&baz=2

Answer 1

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}}

Answer 2

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 }}

See:

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