python-2.7,django-views,django-filter,django-pagination
please make the changes given below in the template, the code below will allow the pagination to get the filtered value and so it will paginate through the filtered value. Do the same changes for the next </table> <span class="step-links"> {% if leads.has_previous %} <a href="?page={{ leads.previous_page_number...
First of all I would change the following: try: blogs = paginator.page(page) except(EmptyPage, InvalidPage): blogs = paginator.page(page) # Raises the same error But you could pass a range within your context. index = paginator.page_range.index(blogs.number) max_index = len(paginator.page_range) start_index = index - 3 if index >= 3 else 0 end_index =...
django,django-rest-framework,django-pagination
When using regular APIView, you need to use Django's own Paginator class. Django Pagination in Views In your case you can paginate queryset before sending it to serializer. Something like this: def get(self, request, format=None): try: cart = request.user.cart except Cart.DoesNotExist: cart = Cart.objects.create(user=request.user) cart_details = cart.cart_details.all() paginator = Paginator(cart_details,...
django,django-class-based-views,django-pagination
The pagination is done in paginate_queryset, called by get_context_data, called by post (that defaults to a subcall to get) , so you can so something like (given you have a filter method that filters based on post data): def post(self, request, *args, **kwargs): self.queryset = self.filter(self.get_queryset()) return super(MyView, self).get(request, *args,...
You should move to Post/Redirect/Get pattern. You need 2 views, first one for form and second for results. Create a new url able to parse all three parameters plus page. /(?P<country>\w+)/(?P<province>\w+)/(?P<site>\w+)/(?P<page>\d+). Route this url to a new view show_results. This view shows page results filtering for all three parameters. In...