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 a full URL like /foo/bar?p

Answer 1

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.