django-urls

How do I set urlpatterns based on domain name or TLD, in Django? For some links, Amazon shows url in native language based on its website tld. http://www.amazon.de/bücher-buch-literatur/ ( de : books => bücher ) http://www.amazon.fr/Nouveautés-paraître-Livres/ ( fr : books => Livres ) http://www.amazon.co.jp/和書-ユーズドブッ-英語学習/ ( jp : books => 和書 ) ( the links are incomplete and just show as samples. ) Is it possible to get host name in urls.py? (request object is not available in urls.py) or maybe in process_request of middleware and use it in urls.py(how???) Any alternate suggestions how to

Is possible to add GET variables in a redirect ? (Without having to modifiy my urls.py) If I do redirect('url-name', x) I get HttpResponseRedirect('/my_long_url/%s/', x) I don't have complains using HttpResponseRedirect('/my_long_url/%s/?q=something', x) instead, but just wondering... Is possible to add GET variables in a redirect ? (Without having to modifiy my urls.py) I don't know of any way to do this without modifying the urls.py . I don't have complains using HttpResponseRedirect('/my_long_url/%s/?q=something', x) instead, but just wondering... You might want to write a thin wrapper to

You have a URL which accepts a first_name and last_name in Django: ('^(?P<first_name>[a-zA-Z]+)/(?P<last_name>[a-zA-Z]+)/$','some_method'), How would you include the OPTIONAL URL token of title , without creating any new lines. What I mean by this is, in an ideal scenario: #A regex constant OP_REGEX = r'THIS IS OPTIONAL<title>[a-z]' #Ideal URL ('^(?P<first_name>[a-zA-Z]+)/(?P<last_name>[a-zA-Z]+)/OP_REGEX/$','some_method'), Is this possible without creating a new line i.e. ('^(?P<first_name>[a-zA-Z]+)/(?P<last_name>[a-zA-Z]+)/(?P<title>[a-zA-Z]+)/$','some_method'), ('^(?P<first_name>[a-zA-Z]+)

I have to build an url dynamically according to the current url. Using the {% url %} tag is the easiest way to do it, but I need the current url name to generate the new one dynamically. How can I get the url name attached to the urlconf that leads to the current view? EDIT : I know I can manually handcraft the url using get_absolute_url but I'd rather avoid it since it's part of a lecture and I would like to demonstrate only one way to build urls. The students know how to use {% url %} . They are know facing a problem when they have to generate a more complete url based on the current one.

Before Django 1.0 there was an easy way to get the admin url of an object, and I had written a small filter that I'd use like this: <a href="{{ object|admin_url }}" .... > ... </a> Basically I was using the url reverse function with the view name being 'django.contrib.admin.views.main.change_stage' reverse( 'django.contrib.admin.views.main.change_stage', args=[app_label, model_name, object_id] ) to get the url. As you might have guessed, I'm trying to update to the latest version of Django, and this is one of the obstacles I came across, that method for getting the admin url doesn't work