Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community!

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

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?

myproject/urls.py
from django.conf.urls import url, include

from . import views

urlpatterns = [
    (r'^$', views.WelcomeView.as_view(), name="WelcomeView"),
]
myproject/views.py
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 91,250 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
MOD
Chris Freeman
Treehouse Moderator 68,404 Points

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

ā€¢ call the url() function with the expressions within the parens

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