How to overcome “datetime.datetime not JSON serializable”?

Question

I have a basic dict as follows:

sample = {}
sample[\'title\'] = \"String\"
sample[\'somedate\'] = somedatetimehere


        

Answer 1

Try this one with an example to parse it:

#!/usr/bin/env python

import datetime
import json

import dateutil.parser  # pip install python-dateutil


class JSONEncoder(json.JSONEncoder):

    def default(self, obj):
        if isinstance(obj, datetime.datetime):
            return obj.isoformat()
        return super(JSONEncoder, self).default(obj)


def test():
    dts = [
        datetime.datetime.now(),
        datetime.datetime.now(datetime.timezone(-datetime.timedelta(hours=4))),
        datetime.datetime.utcnow(),
        datetime.datetime.now(datetime.timezone.utc),
    ]
    for dt in dts:
        dt_isoformat = json.loads(json.dumps(dt, cls=JSONEncoder))
        dt_parsed = dateutil.parser.parse(dt_isoformat)
        assert dt == dt_parsed
        print(f'{dt}, {dt_isoformat}, {dt_parsed}')
        # 2018-07-22 02:22:42.910637, 2018-07-22T02:22:42.910637, 2018-07-22 02:22:42.910637
        # 2018-07-22 02:22:42.910643-04:00, 2018-07-22T02:22:42.910643-04:00, 2018-07-22 02:22:42.910643-04:00
        # 2018-07-22 06:22:42.910645, 2018-07-22T06:22:42.910645, 2018-07-22 06:22:42.910645
        # 2018-07-22 06:22:42.910646+00:00, 2018-07-22T06:22:42.910646+00:00, 2018-07-22 06:22:42.910646+00:00


if __name__ == '__main__':
    test()

Answer 2

Updated for 2018

The original answer accommodated the way MongoDB "date" fields were represented as:

{"$date": 1506816000000}

If you want a generic Python solution for serializing datetime to json, check out @jjmontes' answer for a quick solution which requires no dependencies.

---

As you are using mongoengine (per comments) and pymongo is a dependency, pymongo has built-in utilities to help with json serialization:
http://api.mongodb.org/python/1.10.1/api/bson/json_util.html

Example usage (serialization):

from bson import json_util
import json

json.dumps(anObject, default=json_util.default)

Example usage (deserialization):

json.loads(aJsonString, object_hook=json_util.object_hook)

---

Django

Django provides a native DjangoJSONEncoder serializer that deals with this kind of properly.

See https://docs.djangoproject.com/en/dev/topics/serialization/#djangojsonencoder

from django.core.serializers.json import DjangoJSONEncoder

return json.dumps(
  item,
  sort_keys=True,
  indent=1,
  cls=DjangoJSONEncoder
)

One difference I've noticed between DjangoJSONEncoder and using a custom default like this:

import datetime
import json

def default(o):
    if isinstance(o, (datetime.date, datetime.datetime)):
        return o.isoformat()

return json.dumps(
  item,
  sort_keys=True,
  indent=1,
  default=default
)

Is that Django strips a bit of the data:

 "last_login": "2018-08-03T10:51:42.990", # DjangoJSONEncoder 
 "last_login": "2018-08-03T10:51:42.990239", # default

So, you may need to be careful about that in some cases.

Answer 3

I have an application with a similar issue; my approach was to JSONize the datetime value as a 6-item list (year, month, day, hour, minutes, seconds); you could go to microseconds as a 7-item list, but I had no need to:

class DateTimeEncoder(json.JSONEncoder):
    def default(self, obj):
        if isinstance(obj, datetime.datetime):
            encoded_object = list(obj.timetuple())[0:6]
        else:
            encoded_object =json.JSONEncoder.default(self, obj)
        return encoded_object

sample = {}
sample['title'] = "String"
sample['somedate'] = datetime.datetime.now()

print sample
print json.dumps(sample, cls=DateTimeEncoder)

produces:

{'somedate': datetime.datetime(2013, 8, 1, 16, 22, 45, 890000), 'title': 'String'}
{"somedate": [2013, 8, 1, 16, 22, 45], "title": "String"}

Answer 4

I had encountered same problem when externalizing django model object to dump as JSON. Here is how you can solve it.

def externalize(model_obj):
  keys = model_obj._meta.get_all_field_names() 
  data = {}
  for key in keys:
    if key == 'date_time':
      date_time_obj = getattr(model_obj, key)
      data[key] = date_time_obj.strftime("%A %d. %B %Y")
    else:
      data[key] = getattr(model_obj, key)
  return data

Answer 5

My solution (with less verbosity, I think):

def default(o):
    if type(o) is datetime.date or type(o) is datetime.datetime:
        return o.isoformat()

def jsondumps(o):
    return json.dumps(o, default=default)

Then use jsondumps instead of json.dumps. It will print:

>>> jsondumps({'today': datetime.date.today()})
'{"today": "2013-07-30"}'

I you want, later you can add other special cases to this with a simple twist of the default method. Example:

def default(o):
    if type(o) is datetime.date or type(o) is datetime.datetime:
        return o.isoformat()
    if type(o) is decimal.Decimal:
        return float(o)

Answer 6

Here is my solution:

import json


class DatetimeEncoder(json.JSONEncoder):
    def default(self, obj):
        try:
            return super().default(obj)
        except TypeError:
            return str(obj)

Then you can use it like that:

json.dumps(dictionnary, cls=DatetimeEncoder)