django-urls

I'm currently experimenting with Django and creating apps following the tutorials on the official website. So my urls.py looks like: urlpatterns = patterns('', (r'^/$','ulogin.views.index'), #why doesn't this work? (r'^ucode/$', 'ulogin.views.index'), (r'^ucode/(\d+)/$', 'ulogin.views.index'), ) And my views.py looks like: def index(request): return HttpResponse("Hello, world. You're at the poll index.") def redirect_to_index(request): return HttpResponseRedirect('/ucode/') When I run the server to check the test url, http://127.0.0.1:8000/ucode displays "Hello, world...etc" correctly, and

I have a list of userprofiles that I want to be able to sort and filter. I have been able to do this manually, by manually typing in the URLs, but I haven't been able to code the template page to allow the persistence of a previously-applied filter or sort. Here is the url and template code that I currently have -- # in urls url(r'^talent/filter\:(?P<position>[A-Za-z]+)/sort\:(?P<sort>[A-Za-z]+)$', 'talent_filter', name='talent_filter_sort'), url(r'^talent/filter\:(?P<position>[A-Za-z]+)/$', 'talent_filter', name='talent_filter'), url(r'^talent/sort\:(?P<sort>[A-Za-z]+)/$', 'talent_sort', name

I am from PHP background, and I know that constants can be accessed at most of places in a framework and I think it is same in django as well. I tried to have that URL too in django but I tried to have it from django.contrib. I tried to utilize django's Site class and imported that. But the problem is that at time of loading settings.py I can't import any django contrib. file. So how can I have SITE URL automatically that I can use anywhere, in template as well as at other places.What is the best way to do so? Do any python utility can do so? Whatever you define in your settings.py, for

I have a django url: path('question/<slug:question_slug>/add_vote/', views.AddVoteQuestionView.as_view()) It work fine with english slug but when slug is persian something like this: /question/سوال-تست/add_vote/ django url throw 404 Not Found , is there any solution to catch this perisan slug url? EDIT: I'm using django 2.1.5. It work fine with this url: re_path(r'question/(?P<question_slug>[\w-]+)/add_vote/$', views.AddVoteQuestionView.as_view()) This is an addition to Selcuk answer given here to pass such language/unicode characters you have to Write some custom path converter Use re_path()

In my model, I want to use the domain name (HOST) I'm using in my views. In views that'd be doable, thanks to the "request" object. But how do I do this models methods? Which don't use "HttpRequest" objects? Now I'm setting a global value HOST in settings.py and using it, but that's ugly. Also, I don't really want to manage "Sites" (the Sites app) — Is there a way, I can grab the "by default" Site Host name? Thanks a lot for your help! (and sorry for my poor English) If you're calling the model method from a view, you could add a parameter for the request to the model method and include it

I am looking for the best way to pass either of two arguments to the views from the URL, without allowing any additional arguments. For example, with the following URLs: (r'^friends/requests', 'app.views.pendingFriends'), (r'^friends/offers', 'app.views.pendingFriends'), If it's possible to pass the URL to the views, so that pendingFriends knows which URL it was called from, that would work. However, I can't see a way to do this. Instead, I could supply the arguments ( requests or offers ) in the URL, to a single Django view, (r'^friends/(?P<type>\w+', 'app.views.pendingFriends'), The argument

Is there any way I can simply output text (not HTML) from urls.py instead of calling a function (in views.py or otherwise)? urlpatterns = patterns('', url(r'^$', "Hello World"), ) No, definitly. A url must map a pattern to a callable, and this callable must accept a HttpRequest as first argument and return a HttpResponse . The closer thing you could do would be to map the pattern to a lambda (anonymous function), ie: from django.http import HttpResponse urlpatterns = patterns('', url(r'^$', lambda request: HttpResponse("Hello World", content_type="text/plain")), ) but I would definitly not