django building a get_page_url function in model to return list with pagination number -

i have scenario items paginated list of items in model , want create function return page contains item.

for example

class item(models.model):    subject  = models.charfield(max_length=200) 

assuming have 1000 items in table , return 10 per page. how can build method/function in model tell me in page item fall?

please advise?

you add classmethod takes order_by , paginate_by

import itertools  def grouper(n, iterable, fillvalue=none):     args = [iter(iterable)] * n     return itertools.izip_longest(fillvalue=fillvalue, *args)   class item(models.model):     # attrs      @classmethod     def get_pagination_index(cls, self, order_by, paginate_by):         if order_by:             qs = cls.objects.all().order_by(order_by)         else:             qs = cls.objects.all()         pages = grouper(paginate_by, qs)         = 0         page in pages:             += 1             if self in page:                 return          # return item_index = list(qs).index(self) # exact number