Python URL Not Found

Question

I am using this Python tutorial but looks like the author hasn't updated it because the code is using deprecated Python code, specifically:

from django.conf.urls import patterns

    urlpatterns = patterns('',
        (r'^hello-world/$', index),
    )

I looked up Django documentation and other StackOverFlow questions but I am unable to resolve how I can use the current method. This is what I have added to replace the above:

urlpatterns = [
    url(r'^hello-world$', hello),
]

And here is the full hello.py file:

import sys

from django.conf import settings
from django.conf.urls import url
from django.http import HttpResponse
from django.core.management import execute_from_command_line

settings.configure(
    DEBUG=True,
    SECRET_KEY='thisisabadkeybutitwilldo',
    ROOT_URLCONF=sys.modules[__name__],
)

def index(request):
    return HttpResponse('Hello, World')

import views

urlpatterns = [
    url(r'^hello-world$', hello),
]

if __name__ == "__main__":
    execute_from_command_line(sys.argv)

And this is the error message I get from Python:

Using the URLconf defined in main, Django tried these URL patterns, in this order: ^$ [name='hello'] The current URL, hello, didn't match any of these.

What am I doing wrong?

The code you've provided doesn't work at all and it definitely doesn't raise that error. When I make any necessary changes it works fine. — Cory Madden
– Cory Madden, Commented Aug 9, 2017 at 0:54

Cory Madden · Accepted Answer · 2017-08-09 00:55:57Z

1

Your code, as provided, does not work. I tested the code from the linked tutorial, and it works fine in 1.8. Your code has a random import(import views) and an undefined variable(hello). So I made the logical edits to your code and tested it and it's fine in 1.11:

import sys

from django.conf import settings
from django.conf.urls import url
from django.http import HttpResponse
from django.core.management import execute_from_command_line

settings.configure(
    DEBUG=True,
    SECRET_KEY='thisisabadkeybutitwilldo',
    ROOT_URLCONF=sys.modules[__name__],
)

def index(request):
    return HttpResponse('Hello, World')

urlpatterns = [
    url(r'^hello-world$', index),
]

if __name__ == "__main__":
    execute_from_command_line(sys.argv)

answered Aug 9, 2017 at 0:55

Cory Madden

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2 Comments

Captain Kirk Over a year ago

Thanks! Looks like I just need to use 'index' instead of 'hello' on this line: url(r'^hello-world$', hello),

Cory Madden Over a year ago

No problem. Yea, I figured that was the desired name, but changed it to index instead since that was the first one I saw.

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