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
On 5 Ago, 21:02, Peter Herndon <tphern...@gmail.com> wrote:
> On 08/05/2009 12:26 PM, drakkan wrote:> No I'm not serving static file I have
> a very large json file
> > dinamycally generated (based on a db query)
>
> Please post the relevant view and model code, so we can see what's
> happening. Also, if your db query is actually a really large number of
> db queries, and you are running in debug mode (DEBUG = True in
> settings), then Django caches all of your SQL queries to assist with
> debugging -- at the expense of memory and performance.
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