Return JSON object using MongoEngine Pymongo with Django.

"""
Make "utilities.py" file in your "app directory" with this code,
"""
from mongoengine import *
import mongoengine
import simplejson
from pymongo.objectid import ObjectId
from types import ModuleType
from itertools import groupby

def message():
    print "python is here...."

def encode_model(obj):
    if isinstance(obj, (mongoengine.Document, mongoengine.EmbeddedDocument)):
        out = dict(obj._data)
        for k,v in out.items():
            if isinstance(v, ObjectId):
                out[k] = str(v)
    elif isinstance(obj, mongoengine.queryset.QuerySet):
        out = list(obj)
    elif isinstance(obj, types.ModuleType):
        out = None
    elif isinstance(obj, groupby):
        out = [ (g,list(l)) for g,l in obj ]
    elif isinstance(obj, (list,dict)):
        out = obj
    else:
        raise TypeError, "Could not JSON-encode type '%s': %s" % (type(obj), str(obj))
    return out
"""
and use this line " result = simplejson.dumps(base, default=encode_model)   "
Example
"""

from  utilities import *

def read(self,request):
     base=User.objects.all()
     message();
     print base
     result = simplejson.dumps(base, default=encode_model)
     print "result is >>>",result
     #print serializers.serialize("json", base.all())
     return result
"""
Thumbs up !!
"""