[Django]-Returning to page that brought you there Django

EDIT:

It would be better to use next parameter in the request to redirect to the page that bought us to form page instead of using HTTP_REFERER.

Lets say you are on page some_page.html having a link to the MySodaFormView page. Here, pass the request.path as a next parameter. This will be used when redirecting.

<a href='/my/soda/form/page/?next={{request.path}}'>Create new soda</a> 

Then in MySodaFormView when rendering the page, pass the next parameter in the context. This parameter will be passed in form action and will be used when redirecting.

In your soda formview template, specify next parameter in the form action.

<form method="POST" action="/my/soda/form/page/?next={{next_url}}">

Your view would look something like:

class MySodaFormView(FormView):

    def get_context_data(self, **kwargs):
        context = super(MySodaFormView, self).get_context_data(**kwargs)
        context['next_url'] = self.request.GET.get('next') # pass `next` parameter received from previous page to the context 
        return context

    def get_success_url(self):
        next_url = self.request.GET.get('next')
        if next_url:
            return next_url # return next url for redirection
        return other_url # return some other url if next parameter not present

EDIT: The below approach using HTTP_REFERER might not work sometimes as some browsers have passing referer feature turned off or provide the user an option to disable that feature.

To return to the page that bought you there, you can use HTTP_REFERER header present in the HttpRequest.META dictionary.

HttpRequest.META is a standard Python dictionary containing all available HTTP headers. One of the headers among them is HTTP_REFERER which contains the referring page if any.

Since you are using FormView, you can override the get_success_url() function to redirect on success to the page which bought the user to MySodaFormView. We will get this page using the value of HTTP_REFERER in the request.META dictionary.

from django.views.generic.edit import FormView

class MySodaFormView(FormView):

    def get_success_url(self):
        referer_url = self.request.META.get('HTTP_REFERER') # get the referer url from request's 'META' dictionary
        if referer_url:
            return referer_url # return referer url for redirection
        return other_url # return some other url if referer url not present

Note: Using HTTP_REFERER from request.META dictionary may not be a “best practice” as some browsers have passing referer feature turned off or provide the user an option to disable that feature. In that case, your redirection would not work properly. You could instead pass a ?next= parameter in the url and in your get_success_url() function, use the value of next to get the url to redirect to.

def soda_view(request):
   # your code goes here

   url = "{0}/{1}".format(request.META.get('HTTP_REFERER', '/'), args)

   return HttpResponseRedirect(url)

FormView, along with other generic class-based views that have FormMixin, has a method get_success_url() that you can use to return to the same page. It would look something like this:

from django.core.urlresolvers import reverse

def get_success_url(self):
    return reverse('url_name_of_page_1')

Or, to combine this with Geo Jacob’s answer, get the referring URL out of the HTTP headers:

def get_success_url(self):
    if 'HTTP_REFERER' in request.META:
        return request.META['HTTP_REFERER']
    else:
        # Do something with the error case

As you are using FormView, simply do this:

from django.shortcuts import reverse

class YourView(FormView):
    success_url = reverse('first-page')

In your urls.py:

url(r'/foo/bar/', some.view, name='first-page'),

first-page being the name of the view that renders Page 1 in your diagram.