On Wed, Nov 2, 2011 at 8:25 AM, Thomas Guettler <h...@tbz-pariv.de> wrote:
> # This is my current solution > if get_special_objects().filter(pk=obj.pk).count(): > # yes, it is special > > I can't speak to the "why" of this situation; it seems to me that this could always be converted into a more efficient database query without any unexpected side-effects (and if I really wanted the side effects, I would just write "if obj in list(qs)" instead). In this case, though, I would usually write something like this: if get_special_objects().filter(pk=obj.pk).*exists*(): # yes, it is special I believe that in some cases, the exists() query can be optimized to return faster than a count() aggregation, and I think that the intent of the code appears more clearly. Ian -- Regards, Ian Clelland <clell...@gmail.com> -- 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.
