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 urlAlso 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})