If you're dealing with a list of model instances the best you can do is using serializers.serialize(), it gonna fit your need perfectly.

However, you are to face an issue with trying to serialize a single object, not a list of objects. That way, in order to get rid of different hacks, just use Django's model_to_dict (if I'm not mistaken, serializers.serialize() relies on it, too):

from django.forms.models import model_to_dict

# assuming obj is your model instance
dict_obj = model_to_dict( obj )

You now just need one straight json.dumps call to serialize it to json:

import json
serialized = json.dumps(dict_obj)

That's it! :)

To avoid the array wrapper, remove it before you return the response:

import json
from django.core import serializers

def getObject(request, id):
    obj = MyModel.objects.get(pk=id)
    data = serializers.serialize('json', [obj,])
    struct = json.loads(data)
    data = json.dumps(struct[0])
    return HttpResponse(data, mimetype='application/json')

I found this interesting post on the subject too:

http://timsaylor.com/convert-django-model-instances-to-dictionaries

It uses django.forms.models.model_to_dict, which looks like the perfect tool for the job.

There is a good answer for this and I'm surprised it hasn't been mentioned. With a few lines you can handle dates, models, and everything else.

Make a custom encoder that can handle models:

from django.forms import model_to_dict
from django.core.serializers.json import DjangoJSONEncoder
from django.db.models import Model

class ExtendedEncoder(DjangoJSONEncoder):

    def default(self, o):

        if isinstance(o, Model):
            return model_to_dict(o)

        return super().default(o)

Now use it when you use json.dumps. The superclass DjangoJSONEncoder can handle dates, datetime, timedelta, Decimal, Promise, and UUID objects which aren't natively supported by json.dumps.

json.dumps(data, cls=ExtendedEncoder)

Now models, dates and everything can be serialized and it doesn't have to be in an array or serialized and unserialized. Anything you have that is custom can just be added to the default method.

You can even use Django's native JsonResponse this way:

from django.http import JsonResponse

JsonResponse(data, encoder=ExtendedEncoder)

If you're asking how to serialize a single object from a model and you know you're only going to get one object in the queryset (for instance, using objects.get), then use something like:

import django.core.serializers
import django.http
import models

def jsonExample(request,poll_id):
    s = django.core.serializers.serialize('json',[models.Poll.objects.get(id=poll_id)])
    # s is a string with [] around it, so strip them off
    o=s.strip("[]")
    return django.http.HttpResponse(o, mimetype="application/json")

which would get you something of the form:

{"pk": 1, "model": "polls.poll", "fields": {"pub_date": "2013-06-27T02:29:38.284Z", "question": "What's up?"}}

If you want to return the single model object as a json response to a client, you can do this simple solution:

from django.forms.models import model_to_dict
from django.http import JsonResponse

movie = Movie.objects.get(pk=1)
return JsonResponse(model_to_dict(movie))

It sounds like what you're asking about involves serializing the data structure of a Django model instance for interoperability. The other posters are correct: if you wanted the serialized form to be used with a python application that can query the database via Django's api, then you would wan to serialize a queryset with one object. If, on the other hand, what you need is a way to re-inflate the model instance somewhere else without touching the database or without using Django, then you have a little bit of work to do.

Here's what I do:

First, I use demjson for the conversion. It happened to be what I found first, but it might not be the best. My implementation depends on one of its features, but there should be similar ways with other converters.

Second, implement a json_equivalent method on all models that you might need serialized. This is a magic method for demjson, but it's probably something you're going to want to think about no matter what implementation you choose. The idea is that you return an object that is directly convertible to json (i.e. an array or dictionary). If you really want to do this automatically:

def json_equivalent(self):
    dictionary = {}
    for field in self._meta.get_all_field_names()
        dictionary[field] = self.__getattribute__(field)
    return dictionary

This will not be helpful to you unless you have a completely flat data structure (no ForeignKeys, only numbers and strings in the database, etc.). Otherwise, you should seriously think about the right way to implement this method.

Third, call demjson.JSON.encode(instance) and you have what you want.

.values() is what I needed to convert a model instance to JSON.

.values() documentation: https://docs.djangoproject.com/en/3.0/ref/models/querysets/#values

Example usage with a model called Project.

Note: I'm using Django Rest Framework

from django.http import JsonResponse

@csrf_exempt
@api_view(["GET"])
def get_project(request):
    id = request.query_params['id']
    data = Project.objects.filter(id=id).values()
    if len(data) == 0:
        return JsonResponse(status=404, data={'message': 'Project with id {} not found.'.format(id)})
    return JsonResponse(data[0])

Result from a valid id:

{
    "id": 47,
    "title": "Project Name",
    "description": "",
    "created_at": "2020-01-21T18:13:49.693Z",
}

Here's my solution for this, which allows you to easily customize the JSON as well as organize related records

Firstly implement a method on the model. I call is json but you can call it whatever you like, e.g.:

class Car(Model):
    ...
    def json(self):
        return {
            'manufacturer': self.manufacturer.name,
            'model': self.model,
            'colors': [color.json for color in self.colors.all()],
        }

Then in the view I do:

data = [car.json for car in Car.objects.all()]
return HttpResponse(json.dumps(data), content_type='application/json; charset=UTF-8', status=status)

I solved this problem by adding a serialization method to my model:

def toJSON(self):
    import simplejson
    return simplejson.dumps(dict([(attr, getattr(self, attr)) for attr in [f.name for f in self._meta.fields]]))

Here's the verbose equivalent for those averse to one-liners:

def toJSON(self):
    fields = []
    for field in self._meta.fields:
        fields.append(field.name)

    d = {}
    for attr in fields:
        d[attr] = getattr(self, attr)

    import simplejson
    return simplejson.dumps(d)

_meta.fields is an ordered list of model fields which can be accessed from instances and from the model itself.

Use Django Serializer with python format,

from django.core import serializers

qs = SomeModel.objects.all()
serialized_obj = serializers.serialize('python', qs)

What's difference between json and python format?

The json format will return the result as str whereas python will return the result in either list or OrderedDict

Use list, it will solve problem

Step1:

 result=YOUR_MODELE_NAME.objects.values('PROP1','PROP2').all();

Step2:

 result=list(result)  #after getting data from model convert result to list

Step3:

 return HttpResponse(json.dumps(result), content_type = "application/json")

To serialize and deserialze, use the following:

from django.core import serializers

serial = serializers.serialize("json", [obj])
...
# .next() pulls the first object out of the generator
# .object retrieves django object the object from the DeserializedObject
obj = next(serializers.deserialize("json", serial)).object

All of these answers were a little hacky compared to what I would expect from a framework, the simplest method, I think by far, if you are using the rest framework:

rep = YourSerializerClass().to_representation(your_instance)
json.dumps(rep)

This uses the Serializer directly, respecting the fields you've defined on it, as well as any associations, etc.

It doesn't seem you can serialize an instance, you'd have to serialize a QuerySet of one object.

from django.core import serializers
from models import *

def getUser(request):
    return HttpResponse(json(Users.objects.filter(id=88)))

I run out of the svn release of django, so this may not be in earlier versions.

ville = UneVille.objects.get(nom='lihlihlihlih')
....
blablablab
.......

return HttpResponse(simplejson.dumps(ville.__dict__))

I return the dict of my instance

so it return something like {'field1':value,"field2":value,....}

how about this way:

def ins2dic(obj):
    SubDic = obj.__dict__
    del SubDic['id']
    del SubDic['_state']
return SubDic

or exclude anything you don't want.

Might have missed this, but this seems cleanest:

>>> import json
>>> from django.forms.models import model_to_dict

>>> json.dumps(model_to_dict(my_instance))
'{"code": "AG", "ordering": null, "aliases": ["types"], ...

(aliases above is a successfully serialized JSONField.)

If it is going straight into a response, as @flowfelis suggests use django.http.JsonResponse(model_to_dict(my_instance)).

This is a project that it can serialize(JSON base now) all data in your model and put them to a specific directory automatically and then it can deserialize it whenever you want... I've personally serialized thousand records with this script and then load all of them back to another database without any losing data.

Anyone that would be interested in opensource projects can contribute this project and add more feature to it.

serializer_deserializer_model

Let this is a serializers for CROPS, Do like below. It works for me, Definitely It will work for you also.

First import serializers

from django.core import serializers

Then you can write like this

class CropVarietySerializer(serializers.Serializer): 
    crop_variety_info = serializers.serialize('json', [ obj, ])

OR you can write like this

class CropVarietySerializer(serializers.Serializer): 
    crop_variety_info = serializers.JSONField() 

Then Call this serializer inside your views.py For more details, Please visit https://docs.djangoproject.com/en/4.1/topics/serialization/ serializers.JSONField(*args, **kwargs) and serializers.JSONField() are same. you can also visit https://www.django-rest-framework.org/api-guide/fields/ for JSONField() details.

you can use something like this if you have nested objects as well.

from django.forms import model_to_dict
from django.core.serializers.json import DjangoJSONEncoder
from django.db.models import Model, QuerySet
class ExtendedEncoder(DjangoJSONEncoder):

    def default(self, o):
        if isinstance(o, Model):
            return model_to_dict(o)
        if isinstance(o, QuerySet):
            return [obj for obj in o]

        return super().default(o)

You can use a custom function to achieve that.

def model_to_dict_custom(instance):
model_dict = {}
# Iterate through the model instance's fields
for field in instance._meta.fields:
    field_name = field.name
    field_value = getattr(instance, field_name)
    # Convert special types if needed (e.g., datetime to string)
    if hasattr(field_value, 'strftime'):
        field_value = field_value.strftime('%Y-%m-%d %H:%M:%S')
    model_dict[field_name] = field_value
return model_dict

from django.core import serializers as core_serializers

seedproducerob=commoncertificate.objects.filter(golobalcat=bussnesscatvalue).order_by('-publisheddate')

data = core_serializers.serialize('python', seedproducerob)

return JsonResponse(data,status=200,safe=False)