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I am having trouble to open a web page at the specific position from django view. After submitting a form, I wish it display where it was, but I cannot know where it has to be on the template itself, since that information is in form values.

lionlinekz
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  • Hi and welcome on StackOverflow! :) Add your code, framework versions and all related information to help us to understand your problem. – Louis Barranqueiro Dec 29 '15 at 13:37
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    Are you looking for this: http://stackoverflow.com/questions/16388772/maintain-scroll-position-of-large-html-page-when-client-returns ? – karthikr Dec 29 '15 at 14:06

1 Answers1

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Sounds like you need to define your success url as the same page from where the form was submitted, plus an anchor.

For example:

from os.path import join

class MyFormView(FormView):
    template_name='some_form_template.html'

    def get_success_url(self):
        """
        Returns the supplied success URL.
        """
        return join(reverse('viewname'), '#anchor')

In your some_form_template.html template:

<a id="anchor">Where you'd like the user to be dropped on the page.</a>

If your form validation was successful, the page will refresh and the user will be dropped to the anchor specified on that template.

Patrick Beeson
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