On Wed, 2009-08-05 at 13:09 -0700, drakkan wrote:
> The code is something like this
>
> obj=DBModels.objects.filter(somefilter)
> results=[]
> for o in obj:
> results.append({'field1':o.field1,'field2':o.field2})
>
> return JsonResponse(results)
>
> where JsonResponse is a class derived from HttpResponse that add a
> simplejson.dumps
>
> I have a really large database and in some edge case I need to return
> a large number of results (about 150.000 when I get the error). I make
> only one query with select_related and I tried also with DEBUG=False
> but nothing changed only the detailed stack trace isn't printed,
>
> thanks
> drakkan
>
>
Then don't build up a big list, use and consume:
def worker():
yield '['
first = True
for o in DBModels.objects.filter(somefilter).iterator():
if not first:
yield ','
else:
first = False
yield simplejson.dumps({'field1':o.field1,'field2':o.field2})
yield ']'
return HttpResponse(worker(), mimetype='application/json')
PS - AJAX applications aren't nice and snappy when dealing with that
quantity of information...
Tom
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