On Wed, Nov 2, 2011 at 8:25 AM, Thomas Guettler <hv@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
<clelland@gmail.com>
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