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 you programatically do the same in Python code?

What I need is to create a list of menu items where each item has name, URL, and an active flag (whether it's the current page or not). This is because it will be a lot cleaner to do this in Python than the template language.

1
28
10/14/2013 2:30:01 PM

Accepted Answer

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)

https://docs.djangoproject.com/en/dev/ref/urlresolvers/

34
5/6/2015 3:39:18 PM

I'm using two different approaches in my models.py. The first is the permalink decorator:

from django.db.models import permalink

def get_absolute_url(self): 
    """Construct the absolute URL for this Item."""
    return ('project.app.views.view_name', [str(self.id)])
get_absolute_url = permalink(get_absolute_url)

You can also call reverse directly:

from django.core.urlresolvers import reverse

def get_absolute_url(self): 
    """Construct the absolute URL for this Item."""
    return reverse('project.app.views.view_name', None, [str(self.id)])

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