Trigger a modal on django form submit












0















I have a series of Django forms and formsets that are connected together in a series of steps. When the final form step is submitted a celery task is started to create an entity based on all the information provided. The way this is currently working, is that the final form is submitted (and the task starts) and then redirects to a page which lets the user know that the entity is being created and then redirects once this is complete. However, I now want to have this loading page show up as a modal on the previous page, rather than show on an entirely new page.



I have previously been doing this almost solely in Django.



When I submit the final form the form_valid looks like this (I am using django-extra-views to make formsets so it is slighlty different from normal):



def forms_valid(self, form, inlines):
self.object = form.save()

models.ActivityBaseNext.objects.create(
activity_base=self.activity_base,
order=self.activity_base.get_next_next_step_order(),
activity_configure=self.object,
)

for formset in inlines:
formset.save()

transaction.on_commit(
tasks.activity_configure.s(
id=self.object.pk,
).delay
)

return HttpResponseRedirect(self.object.get_current_next_url())


which causes the redirect to the loading page, which looks like so:



class ActivityConfigureStatus(LoginRequiredMixin, DetailView):
model = models.Activity
template_name = 'alert.html'

title = 'Create Activity'

display_text = 'Importing activity data'

refresh_secs = 1

def get(self, request, *args, **kwargs):
self.object = get_object_or_404(models.Activity, pk=self.kwargs['pk'])

object_url = self.object.get_current_next_url()

if object_url == reverse('activities:activity_configure_status', kwargs={'pk': self.object.pk}):
context = self.get_context_data(object=self.object)
return self.render_to_response(context)
else:
return HttpResponseRedirect(object_url)


I'm trying to turn the ActivityConfigureStatus view into a modal which appears in the previous view. Ideally I would like to change the code as little so what I am currently trying to do is populate the modal with the relevant data when the submit button is pressed.



I have written the submit button as:



<button type="button" class="btn btn-primary float-right" data-toggle="modal" data-target="#configureModal">Submit</button>


But I am struggling with how to populate the modal with the ActivityConfigureStatus view.



Can someone help me update my code so that the modal content will appear.



Any help would be much appreciated.



Thanks for your time.










share|improve this question





























    0















    I have a series of Django forms and formsets that are connected together in a series of steps. When the final form step is submitted a celery task is started to create an entity based on all the information provided. The way this is currently working, is that the final form is submitted (and the task starts) and then redirects to a page which lets the user know that the entity is being created and then redirects once this is complete. However, I now want to have this loading page show up as a modal on the previous page, rather than show on an entirely new page.



    I have previously been doing this almost solely in Django.



    When I submit the final form the form_valid looks like this (I am using django-extra-views to make formsets so it is slighlty different from normal):



    def forms_valid(self, form, inlines):
    self.object = form.save()

    models.ActivityBaseNext.objects.create(
    activity_base=self.activity_base,
    order=self.activity_base.get_next_next_step_order(),
    activity_configure=self.object,
    )

    for formset in inlines:
    formset.save()

    transaction.on_commit(
    tasks.activity_configure.s(
    id=self.object.pk,
    ).delay
    )

    return HttpResponseRedirect(self.object.get_current_next_url())


    which causes the redirect to the loading page, which looks like so:



    class ActivityConfigureStatus(LoginRequiredMixin, DetailView):
    model = models.Activity
    template_name = 'alert.html'

    title = 'Create Activity'

    display_text = 'Importing activity data'

    refresh_secs = 1

    def get(self, request, *args, **kwargs):
    self.object = get_object_or_404(models.Activity, pk=self.kwargs['pk'])

    object_url = self.object.get_current_next_url()

    if object_url == reverse('activities:activity_configure_status', kwargs={'pk': self.object.pk}):
    context = self.get_context_data(object=self.object)
    return self.render_to_response(context)
    else:
    return HttpResponseRedirect(object_url)


    I'm trying to turn the ActivityConfigureStatus view into a modal which appears in the previous view. Ideally I would like to change the code as little so what I am currently trying to do is populate the modal with the relevant data when the submit button is pressed.



    I have written the submit button as:



    <button type="button" class="btn btn-primary float-right" data-toggle="modal" data-target="#configureModal">Submit</button>


    But I am struggling with how to populate the modal with the ActivityConfigureStatus view.



    Can someone help me update my code so that the modal content will appear.



    Any help would be much appreciated.



    Thanks for your time.










    share|improve this question



























      0












      0








      0








      I have a series of Django forms and formsets that are connected together in a series of steps. When the final form step is submitted a celery task is started to create an entity based on all the information provided. The way this is currently working, is that the final form is submitted (and the task starts) and then redirects to a page which lets the user know that the entity is being created and then redirects once this is complete. However, I now want to have this loading page show up as a modal on the previous page, rather than show on an entirely new page.



      I have previously been doing this almost solely in Django.



      When I submit the final form the form_valid looks like this (I am using django-extra-views to make formsets so it is slighlty different from normal):



      def forms_valid(self, form, inlines):
      self.object = form.save()

      models.ActivityBaseNext.objects.create(
      activity_base=self.activity_base,
      order=self.activity_base.get_next_next_step_order(),
      activity_configure=self.object,
      )

      for formset in inlines:
      formset.save()

      transaction.on_commit(
      tasks.activity_configure.s(
      id=self.object.pk,
      ).delay
      )

      return HttpResponseRedirect(self.object.get_current_next_url())


      which causes the redirect to the loading page, which looks like so:



      class ActivityConfigureStatus(LoginRequiredMixin, DetailView):
      model = models.Activity
      template_name = 'alert.html'

      title = 'Create Activity'

      display_text = 'Importing activity data'

      refresh_secs = 1

      def get(self, request, *args, **kwargs):
      self.object = get_object_or_404(models.Activity, pk=self.kwargs['pk'])

      object_url = self.object.get_current_next_url()

      if object_url == reverse('activities:activity_configure_status', kwargs={'pk': self.object.pk}):
      context = self.get_context_data(object=self.object)
      return self.render_to_response(context)
      else:
      return HttpResponseRedirect(object_url)


      I'm trying to turn the ActivityConfigureStatus view into a modal which appears in the previous view. Ideally I would like to change the code as little so what I am currently trying to do is populate the modal with the relevant data when the submit button is pressed.



      I have written the submit button as:



      <button type="button" class="btn btn-primary float-right" data-toggle="modal" data-target="#configureModal">Submit</button>


      But I am struggling with how to populate the modal with the ActivityConfigureStatus view.



      Can someone help me update my code so that the modal content will appear.



      Any help would be much appreciated.



      Thanks for your time.










      share|improve this question
















      I have a series of Django forms and formsets that are connected together in a series of steps. When the final form step is submitted a celery task is started to create an entity based on all the information provided. The way this is currently working, is that the final form is submitted (and the task starts) and then redirects to a page which lets the user know that the entity is being created and then redirects once this is complete. However, I now want to have this loading page show up as a modal on the previous page, rather than show on an entirely new page.



      I have previously been doing this almost solely in Django.



      When I submit the final form the form_valid looks like this (I am using django-extra-views to make formsets so it is slighlty different from normal):



      def forms_valid(self, form, inlines):
      self.object = form.save()

      models.ActivityBaseNext.objects.create(
      activity_base=self.activity_base,
      order=self.activity_base.get_next_next_step_order(),
      activity_configure=self.object,
      )

      for formset in inlines:
      formset.save()

      transaction.on_commit(
      tasks.activity_configure.s(
      id=self.object.pk,
      ).delay
      )

      return HttpResponseRedirect(self.object.get_current_next_url())


      which causes the redirect to the loading page, which looks like so:



      class ActivityConfigureStatus(LoginRequiredMixin, DetailView):
      model = models.Activity
      template_name = 'alert.html'

      title = 'Create Activity'

      display_text = 'Importing activity data'

      refresh_secs = 1

      def get(self, request, *args, **kwargs):
      self.object = get_object_or_404(models.Activity, pk=self.kwargs['pk'])

      object_url = self.object.get_current_next_url()

      if object_url == reverse('activities:activity_configure_status', kwargs={'pk': self.object.pk}):
      context = self.get_context_data(object=self.object)
      return self.render_to_response(context)
      else:
      return HttpResponseRedirect(object_url)


      I'm trying to turn the ActivityConfigureStatus view into a modal which appears in the previous view. Ideally I would like to change the code as little so what I am currently trying to do is populate the modal with the relevant data when the submit button is pressed.



      I have written the submit button as:



      <button type="button" class="btn btn-primary float-right" data-toggle="modal" data-target="#configureModal">Submit</button>


      But I am struggling with how to populate the modal with the ActivityConfigureStatus view.



      Can someone help me update my code so that the modal content will appear.



      Any help would be much appreciated.



      Thanks for your time.







      django bootstrap-modal






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 28 '18 at 11:22







      BeeNag

















      asked Nov 28 '18 at 11:14









      BeeNagBeeNag

      33231128




      33231128
























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