Here is my code:
urls.py
url(r"magasin/(?P<magasin_id>\d+)/evenement/new/$",
EvenementCreateView.as_view(model=Evenement),
name='new_evenement_magasin'),
views.py:
class EvenementCreateView(CreateView):
form_class = EvenementForm
sucess_url = 'user_magasins'
def dispatch(self, *args, **kwargs):
bla bla bla ....
def form_valid(self, form):
bl bla bla .....
return HttpResponseRedirect(self.get_success_url())
What is funny here is whatever url of the site I call, even one which
doesn't use this view I've the error describe in the subject. The
traceback point on the line of the urls.py.
I'm not python and django proof, but as far I know class blablabla()
define a Class and not a function !!!
Does someone have an idea here ?
Regards
Alain
...
(r"magasin/(?P<magasin_id>\d+)/evenement/new/$",EvenementCreateView.as_view(model=Evenement,)),
...
I haven’t used Class based Views yet, but everytime they show up in the docs it is always without using the url function and giving a name. So this is just a wild guess.
It seems like it is interpreting EvenementCreateView as a view function not as a class and does not let you call as_view.
The error happens always because django looks at all the urls you configured in order to know which one to use, this is why it always crashes.
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url(r"magasin/(?P<magasin_id>\d+)/evenement/new/$",
EvenementCreateView.as_view(model=Evenement),
name='new_evenement_magasin'),
Eugene
I've done the test with another class, it's the same. Then I think @login_required works only for function and not for Class.