On Oct 11, 9:22 am, The Danny Bos <danny...@gmail.com> wrote: > Heya, thanks for all the help of late, I'm about four late nights from > finishing my first big app I'm guessing. > > Anyway, > I'm working on paging in my app and am completely stuck (again), > here's what I'm trying for: > eg:http://www.domainname.com/card/pokemon/0-50/ > > Which would obviously display a list of 'Pokemon' cards from numbers 0 > to 50. > > To get those numbers from the URL, I'm doing: > paging = str(paging) > paging = paging.split("-") > page_from = paging[0] > page_to = paging[1] > items = Cards.objects.all()[page_from:page_to] > > Seems to work but may need some error checking, yeh? > > Where I'm really stuck is listing out the pages in the template like > so: > <a href="">0-50</a>, <a href="">51-100</a>, <a href="">101-132</a>. > > I guess I need to get the Count of the records first, in this case > "132". Then how would I loop through and create those links > automatically. As each Card set may be different obviously, some may > have "43" records, some "682" records etc ... But I still want to page > through by "50" each time using the URL. > > Any help would be great, I'm really stumped on this one. > Hope that's enough info to get some help. > > d
There is a whole section of the documentation devoted to pagination: http://docs.djangoproject.com/en/dev/topics/pagination/ including some useful built-in functionality. -- DR. --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
