success_url in UpdateView, based on passed value
Create a class MyUpdateView
inheritted from UpdateView
and override get_success_url
method:
class MyUpdateView(UpdateView):
def get_success_url(self):
pass #return the appropriate success url
Also i like to pass such parameters like template_name and model inside of inheritted class view, but not in .as_view()
in urls.py
Had the same issue. Was able to get the paramater from self.kwargs as Dima mentioned:
def get_success_url(self):
if 'slug' in self.kwargs:
slug = self.kwargs['slug']
else:
slug = 'demo'
return reverse('app_upload', kwargs={'pk': self._id, 'slug': slug})
I found a way which is useful and very simple. Check it out.
class EmployerUpdateView(UpdateView):
model = Employer
#other stuff.... to be specified
def get_success_url(self):
pk = self.kwargs["pk"]
return reverse("view-employer", kwargs={"pk": pk})