How to 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 a full URL like /foo/bar?param=1&baz=2

Answer 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.

Answer 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.