django-views

Somewhat of a follow-up to the question posted here ( Django: modifying data with user input through custom template tag? ), but since asking the question I've decided to take a different approach. As you can tell I'm a newb so please go easy. I want a weekly calendar in my Django app that displays shifts in the database. Following is the Shift model. class Shift(models.Model): owner = models.ForeignKey(settings.AUTH_USER_MODEL, related_name='shifts', blank=True, null=True) # A shift may not have an owner. # Choices for fields. Add more choices as necessary. DAYS = ( ('M', 'Monday'), ('TU',

I've got a URLs pattern like this: urlpatterns = ( url(r'^$', list_titles, name='list'), url(r'^(?P<tag>[a-z\-0-9]+?)/$', list_titles, name='filtered-list'), url(r'^(?P<title>\S+?)/$', show_title, name='title'), ) The filtered-list and title match the same things. If there is an available list of things matching the tag in filtered-list , I want list_titles to fire off. But if there isn't a matching tag , I want to bubble that back to the URL processor so show_title fires off. If there's no matching title, I'll raise a proper 404 there. I know I can do this from inside the view...but it's a

I know this is a simple question, sorry. I just want to return a simple string, no templates. I have my view: def myview(request): return "return this string" I don't remember the command. Thanks According to the documentation : A view function, or view for short, is simply a Python function that takes a Web request and returns a Web response. Each view function is responsible for returning an HttpResponse object. In other words, your view should return a HttpResponse instance: from django.http import HttpResponse def myview(request): return HttpResponse("return this string") ThePhi You can't

I would like to pass a number to my generic view (DetailView) to get one object Here is my code Urlpattern (r'^newreportview/(?P<number>\w+)/$', NewReportView.as_view()), View Class class NewReportView(DetailView): template_name = "report/newreportview.html" context_object_name = "newreportview" def get_queryset(self): task= get_object_or_404(MyTask,applicationnnumber=self.args[0]) return task I guess something is wrong in this line name = get_object_or_404(MyTask,applicationnnumber=self.args[0]) error message: Exception Type: IndexError Exception Value: tuple index out of range How should I

I want to make request to url and django view should read the file and send the http response back to play the same file in browser.I got the following code but it does't play anything please anyone help me.. Right now i hard coded the file name in the code. url: http://localhost/playfile/audiofile_name def playAudioFile(request): try: fname="C:\\test\\audio\\t.mp3" wrapper = FileWrapper(file(fname)) print content_type response = HttpResponse(wrapper, content_type="audio/mpeg") print response response['Content-Length'] =os.path.getsize(fname ) return response except: return HttpResponse()

I am rewriting a big piece of our application which requires a user to create a Project with Rewards attached to it. The form is broken into different steps, the first two are the normal Project , the next one is the Rewards , and then lastly a simple preview that lets the user flick back and forth to create a perfect Project . my forms.py class BaseRewardFormSet(BaseModelFormSet): def __init__(self, *args, **kwargs): super(BaseRewardFormSet, self).__init__(*args, **kwargs) self.queryset = Reward.objects.none() RewardFormSet1 = modelformset_factory(Reward, extra=2, exclude=('project'), formset