I have a quite simple query set and a related generic views:
f_detail = {
'queryset': Foto.objects.all(),
'template_name': 'foto_dettaglio.html',
"template_object_name" : "foto",
}
urlpatterns = patterns('',
# This very include
(r'^foto/(?P<object_id>\d+)/$', list_detail.object_detail, f_detail, ),
)
Just a template for generating a detail page of a photo: so there's no view.
Is there an easy way to have a link to previous | next element in the template without manualy coding a view ?
Somthing like a:
{% if foto.next_item %}
<a href="/path/foto/{{ foto.next_item.id_field }}/">Next</a>
{% endif}
class Foto(model):
...
def get_next(self):
next = Foto.objects.filter(id__gt=self.id)
if next:
return next.first()
return False
def get_prev(self):
prev = Foto.objects.filter(id__lt=self.id).order_by('-id')
if prev:
return prev.first()
return False
you can tweak these to your liking. i just looked at your question again... to make it easier than having the if statement, you could make the methods return the markup for the link to the next/prev if there is one, otherwise return nothing. then you'd just do foto.get_next
etc. also remember that querysets are lazy so you're not actually getting tons of items in next/prev.
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