How to generate urls in django

Question

In Django's template language, you can use {% url [viewname] [args] %} to generate a URL to a specific view with parameters. How can I perform the same in Python code?

 I need to create a list of menu items where each item has name, URL, and an active flag. This is because it will be a lot cleaner to do this in Python than the template language.

Can anyone tell me how to do this?

Answer 1

If you need to use something similar to the {% url %} template tag in your code, Django provides the django.core.urlresolvers.reverse(). The reverse function has the following signature:

reverse(viewname, urlconf=None, args=None, kwargs=None)