问题
Works like a charm:
MyCreateView(CreateView):
template_name = "my_template_name"
form_class = MyModelForm
success_url = "/success/"
But the following doesn't:
MyUpdateView(UpdateView):
template_name = "my_template_name"
form_class = MyModelForm
success_url = "/success/"
I get this error:
MyUpdateView is missing a queryset. Define MyUpdateView.model, MyUpdateView.queryset, or override MyUpdateView.get_queryset().
Why does an UpdateView
need model
, queryset
or get_queryset
defined to not cause an error while CreateView
doesn't? Shouldn't it be able to automatically derive it from the Model used in the ModelForm?
回答1:
Currently (django 1.5.1 official release) UpdateView
is calling self.get_object()
to be able to provide instance
object to Form.
From https://github.com/django/django/blob/1.5c2/django/views/generic/edit.py#L217:
def get(self, request, *args, **kwargs):
self.object = self.get_object()
return super(BaseUpdateView, self).get(request, *args, **kwargs)
def post(self, request, *args, **kwargs):
self.object = self.get_object()
return super(BaseUpdateView, self).post(request, *args, **kwargs)
And self.get_object
method needs one of this properties declared: model
, queryset
or get_queryset
Whereas CreateView
don't call self.get_object()
.
From https://github.com/django/django/blob/1.5c2/django/views/generic/edit.py#L194:
def get(self, request, *args, **kwargs):
self.object = None
return super(BaseCreateView, self).get(request, *args, **kwargs)
def post(self, request, *args, **kwargs):
self.object = None
return super(BaseCreateView, self).post(request, *args, **kwargs)
回答2:
You might have a problem in your urls.py file.
What I think you wrote in it is:
url(r'foldername/(?P[0-9]+)/$', views.UpdateView.as_view(), name='update'),
but you have to change UpdateView to MyUpdateView, like this:
url(r'foldername/(?P[0-9]+)/$', views.MyUpdateView.as_view(), name='update'),
来源:https://stackoverflow.com/questions/17613398/why-does-updateview-need-to-have-model-queryset-get-queryset-defined-when-using