Reputation: 3252
I have a Django model containing DecimalField. The resulting json should contain only data (no keys) so I'm using values_list() to convert queryset to list of Tuples:
MyModel.objects.filter(...).values_list('my_date_field','my_decimal_field').order_by('my_date_field')
Then, I need to serialize it to json... but json.dumps does not seems to be able to process the Decimal field... A lot of SO answers about that suggest to make your own encoder to use with json.dumps but those custom encoders are not recursive and seems not to work with a list of Tuple...
What I need is returning json with this format:
[[1162512000000,78.29],
[1162771200000,79.71],
[1162857600000,80.51],
[1162944000000,82.45],
[1163030400000,83.34],
[1163116800000,83.12],
[1163376000000,84.35]]
It seems to me that this should be a simple task but can't find a simple way to do it without having to parse and process everything manually...
Any suggestions?
Thanks a lot
Etienne
Upvotes: 2
Views: 880
Reputation: 3174
This should work:
import json
from decimal import Decimal as D
class DecimalJSONEncoder(json.JSONEncoder):
def default(self, o):
if type(o) == D:
# Here You can decide if You want decimal to be converted
# to string or float.
return float(o)
return super(DecimalJSONEncoder, self).default(o)
data = [[1162512000000, D(78.29)],
[1162771200000, D(79.71)],
[1162857600000, D(80.51)],
[1162944000000, D(82.45)],
[1163030400000, D(83.34)],
[1163116800000, D(83.12)],
[1163376000000, D(84.35)]]
encoder = DecimalJSONEncoder()
encoder.encode(data)
# Result:
# '[[1162512000000, 78.29], [1162771200000, 79.71], [1162857600000, 80.51], ...'
Upvotes: 1