Welcome to the Treehouse Community

The Treehouse Community is a meeting place for developers, designers, and programmers of all backgrounds and skill levels to get support. Collaborate here on code errors or bugs that you need feedback on, or asking for an extra set of eyes on your latest project. Join thousands of Treehouse students and alumni in the community today. (Note: Only Treehouse students can comment or ask questions, but non-students are welcome to browse our conversations.)

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today.

Python Django Class-based Views Classy Views Basic View

Didn't find an URL for your view

I am trying to create an URL for my class-based view, which has a get method, returns an HttpResponse, and is instantiated using the as_view() method in the URL pattern list. I can't see why the error "Didn't find an URL for your view" is happening?

from django.conf.urls import url, include

from . import views

urlpatterns = [
    (r'^$', views.WelcomeView.as_view(), name="WelcomeView"),
from django.views.generic import View
from django.http import HttpResponse

class WelcomeView(View):
    def get(self, request):
        return HttpResponse("hi")
Jonathan Grieve
Jonathan Grieve
Treehouse Moderator 90,705 Points

Unfortunately, I'm not familiar with Django, so I've asked around a few more people for you.

The best thing I can offer is check indentation in your code and your imports are correct.

Try running the code again? Sometimes some funny things happen with the code engine behind the scenes. ;)

1 Answer

Chris Freeman
Chris Freeman
Treehouse Moderator 67,986 Points

You are VERY close. The url patterns should be url objects.

• call the url() function with the expressions within the parens

Post back if you need more help. Good luck!!!