success_url in UpdateView, based on passed value

Question

How can I set success_url based on a parameter?
I really want to go back to where I came from, not some static place. In pseudo code:

url(r         


        

Answer 1

Define get_absolute_url(self) on your model. Example

class Poll(models.Model):
    question = models.CharField(max_length=100)
    slug = models.SlugField(max_length=50)
    # etc ...

    def get_absolute_url(self):
        return reverse('poll', args=[self.slug])

If your PollUpdateView(UpdateView) loads an instance of that model as object, it will by default look for a get_absolute_url() method to figure out where to redirect to after the POST. Then

url(r'^polls/(?P<slug>\w+)/, UpdateView.as_view(
    model=Poll, template_name='generic_form_popup.html'),

should do.

Answer 2

Why don't you add a 'next' parameter to your form (template) and catch it in your view. It's common practice to achieve redirecting this way.

Answer 3

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

Answer 4

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})

Answer 5

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})