I was able to get this to work. There was an import error from my
side,
data =
serializers.serialize('json',Food.objects.filter(group__exact=my_group),fields=('food_name','group','description'))
following two serialization ways gave me exception -
'dict' object has no attribute '_meta'
a) json_out_list= serialize('json',out_list)
json_serializer = serializers.get_serializer("json")()
json_out_list = json_serializer.serialize(out_list,
ensure_ascii=False)
so I am unblocked for now. Thanks everyone
thanks
Ashish Gupta
On Mar 31, 11:49 am, ydjango <[EMAIL PROTECTED]> wrote:
> I am on svn version as of two days back.
>
> I am using values(*fields) to create an output from my table.
> (following using print statement)
>
> out_list=Food.objects.filter(group__exact=my_group).values('food_name','group','description')
> print out_list
>
> [{'food_name': u'nirvana', 'group': 1L, 'description': u'nirvana1'},
> {'food_name': u'dj new versio', 'group': 1L, 'description':
> u'jsklajdkls'}, {'food_name': u'PMC', 'group': 1L, 'description':
> u'jsklajdkls'}]
>
> ( why do I see u before text in output above?)
>
> when I try json serailize It throws an exception:
>
> i tried following, none seems to work, any clues-
>
> from django.core.serializers import serialize
> a) json_out_list= serialize('json',out_list)
> b) json_serializer = serializers.get_serializer("json")()
> json_out_list = json_serializer.serialize(out_list,
> ensure_ascii=False, stream=response)
> c) data = serializers.serialize('json',
> Food.objects.filter(group__exact=my_group),
> fields=('food_name','group','description'))
>
> Any clues?
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