Hi users and devs In the next example for pagination:
contact_list = Contacts.objects.all() paginator = Paginator(contact_list, 25) # Show 25 contacts per page Consider that Contacts.objects.all() returns a list with 1.000.000 objects, and I want to paginate to 25 objects per page. Wouldn't be significantly more fast to query for Contacts.objects.all() [offset:limit]? Instead of cut a big list and querying for all objects at the same time? Regards, Rodrigo --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---