CODEWITHSUNDEEP
pythondjango
Programmer
Programmer

Reputation: 8727

My First Python Web App

I am trying to write a program by running Django using manage.py runserver ip:port in my Linux box as a non-root user. My first objective is that if a user enters a URL in the browser http://ip:port it should display something or a welcome content.

So I modified my mysite/url.py something like this:

urlpatterns = patterns('',
    ('', ''),
    (r'^admin/', include(admin.site.urls)),
)

python manage.py runserver : runs and start fine but in browser I get below exception:

Exception Type: ImproperlyConfigured Exception Value: Empty URL pattern view name not permitted (for pattern '')

So how can construct the URL just like:

www.google.com
www.mail.yahoo.com
www.mysite.com

Thanks in advance

Upvotes: 4

Views: 516

Answers (2)

Edwin
Edwin

Reputation: 2114

Unless you're willing to get down to the nitty-gritty and configure, run, and host your own Apache Server (or another server config) AND register a domain name, you CANNOT change the domain name from within Python/Django.

Django only handles everything that comes after the domain name, such as the /home/ portion of the URL http://www.mysite.com/home/.

If you're up to it, the Django book has a pretty good section on deploying Django and setting up the server to run it, but it doesn't cover all the bases.

Upvotes: 3

Platinum Azure
Platinum Azure

Reputation: 46233

The empty pattern will match anything, so you should be anchoring it to full empty string. The second tuple parameter must also be a view name.

In your URLconf:

urlpatterns = patterns('',
    (r'^$', 'views.home'),     # ^$ means, beginning of string followed by end of string, in other words match on exactly empty string and nothing else
    (r'^admin/', include(admin.site.urls)),
)

In a views.py in the same directory:

from django.http import HttpResponse

def home(request):
    return HttpResponse('Hello World')

Upvotes: 4

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