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Asked: May 11, 20262026-05-11T09:48:38+00:00 2026-05-11T09:48:38+00:00

I’m starting a new web app project using Django and Pinax. I want to

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I’m starting a new web app project using Django and Pinax. I want to be able to give my users unique domain names like WordPress and other sites do : username.wordpress.com. I’m not sure how to approach this with Django, since the url parsing logic (in urls.py) starts with the url AFTER the domain name.

More specifically, there will be multiple groups of users, each group having a unique name. Not sure that makes a difference, but I thought I should mention that.

Is there some way I can manipulate the http request so that the URL looks to Django as if the url were something like www.domain.com/groupname, but still showed in the browser address bar as groupname.domain.com?

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  1. 2026-05-11T09:48:38+00:00Added an answer on May 11, 2026 at 9:48 am

    You can use some custom middleware to intercept the request and get the subdomain from it. The following code will retrieve the subdomain and redirect to a view by reversing the named url.

    Put it in a middleware.py file in your app.

    Make sure you set up the middleware in your settings.py file.

    Make sure you’ve named your view in urls.py

    middleware.py

    from django.http import HttpResponseRedirect from django.core.urlresolvers import reverse import re  subdomain_pattern = re.compile('(?P<subdomain>.*?)\..*?')  class SubdomainMiddleware(object):     def process_request(self, request):         match = subdomain_pattern.match(request.get_host())         subdomain = match.group('subdomain')         redirect_url = reverse('groups_detail', args=[subdomain])         return HttpResponseRedirect(redirect_url) 

    urls.py

    from django.conf.urls.defaults import *  urlpatterns = patterns('',     url(r'^groups/(?P<name>.+)/$', 'groups.views.detail', {}, name='group_detail'), ) 

    Note: this code is untested.

    Redirecting can alter the URL’s appearance. If you want to avoid this, simply call the associated view, capture its result, and return it in an HttpResponse().

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