Create JSON Response in Django with Model

Question

I am having some issue here. I am trying to return a JSON response made of a message and a model instance:

   class MachineModel(models.Model):
       name = models.CharField(max_length=64, blank=False)
       description = models.CharField(max_length=64, blank=False)
       manufacturer = models.ForeignKey(Manufacturer)
       added_by = models.ForeignKey(User, related_name='%(app_label)s_%(class)s_added_by')
       creation_date = models.DateTimeField(auto_now_add=True)
       last_modified = models.DateTimeField(auto_now=True)

    machine_model_model = form.save(commit=False)
    r_user = request.user.userprofile
    machine_model_model.manufacturer_id = manuf_id
    machine_model_model.added_by_id = request.user.id
    machine_model_model.save()
    alert_message = " The'%s' model " % machine_model_model.name
    alert_message += ("for '%s' " % machine_model_model.manufacturer)
    alert_message += "was was successfully created!"
    test = simplejson.dumps(list(machine_model_model))
    data = [{'message': alert_message, 'model': test}]
    response = JSONResponse(data, {}, 'application/json')


class JSONResponse(HttpResponse):
"""JSON response class."""
    def __init__(self, obj='', json_opts={}, mimetype="application/json", *args, **kwargs):
        content = simplejson.dumps(obj, **json_opts)
        super(JSONResponse,self).__init__(content, mimetype, *args, **kwargs)

But I keep getting:

File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/encoder.py", line 178, in default
raise TypeError(repr(o) + " is not JSON serializable")

TypeError: <MachineModel: "Test12"> is not JSON serializable

Why is that? I have seen before:

models = Model.objects.filter(manufacturer_id=m_id)
json = simplejson.dumps(models)

and that works... what is the difference?!

Thanks!

Serhii Holinei · Accepted Answer · 2016-05-03 12:51:03Z

20

You should use django serializers instead of simplejson:

For example, this returns correctly serialized data:

from django.core import serializers
# serialize queryset
serialized_queryset = serializers.serialize('json', some_queryset)
# serialize object
serialized_object = serializers.serialize('json', [some_object,])

edited May 3, 2016 at 12:51

answered Sep 23, 2012 at 16:16

Serhii Holinei

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4 Comments

abisson Over a year ago

I don't really get a JSON object... I get "[{"pk": 86, "model": "machine_models.machinemodel", "fields": {"name": "MX10", "description": "lol", "creation_date": "2012-09-23T16:50:35.709Z", "last_modified": "2012-09-23T16:50:35.709Z", "added_by": 2, "manufacturer": "1"}}]"

Serhii Holinei Over a year ago

It is actualy JSON object. You can deserialize it on a template. By jquery, for example: jQuery.parseJSON(response), which returns a javascript dictionary.

abisson Over a year ago

Ok good it works! But I always get an array (in JQuery and I have to take the first [0]... Is that normal?! It is kind of annoying as well. Would you recommend a REST Api?!

Serhii Holinei Over a year ago

As the second argument of serialize method can be any iterator that yields Django model instances (taken from docs), I dont know any other method to handle with single model instance serialization. Also, checkout this relative question. About REST Api - sorry, I am not competent in this area. Btw, I use the same descriptions in my projects :-)

Community · Accepted Answer · 2020-06-20 09:12:55Z

17

Method-1: Use Django's `python` serializer

I think this answer won't return a JSON or a Python dict/list object. So, use the format python instead of json

from django.core import serializers
# serialize queryset
serialized_queryset = serializers.serialize('python', some_queryset)
# serialize object
serialized_object = serializers.serialize('python', [some_object,])

Django shell response

In [2]: from django.core import serializers                                                                                                                             

In [3]: qs = SomeModel.objects.all()                                                                                                                                    

In [4]: json_res = serializers.serialize('json',qs)                                                                                                                     

In [5]: type(json_res)                                                                                                                                                  
Out[5]: str

In [6]: python_res = serializers.serialize('python',qs)                                                                                                                 

In [7]: type(python_res)                                                                                                                                                
Out[7]: list

#views.py
from django.core import serializers
from django.http.response import JsonResponse


def some_view(request):
    some_queryset = SomeModel.objects.all()
    serialized_queryset = serializers.serialize('python', some_queryset)
    return JsonResponse(serialized_queryset, safe=False)

Method-2: Use Django's `values()` method

The direct use of values() method will throw TypeError exception, so convert the QuerySet to a python list as below,

from django.http.response import JsonResponse


def sample_view(request):
    return JsonResponse(list(SomeModel.objects.all().values()), safe=False)

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered May 7, 2019 at 14:30

JPG

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4 Comments

JPG Over a year ago

@Radesh What happened with method-1?

Radesh Over a year ago

return json response contain some extra /" like this : [{\"model\": \"api\"

Ulvi Over a year ago

@JPG, is it equally the same to use either of these methods? Do any of these have any drawbacks in performance or another thing? I personally see using second method very easy but also curious about if it has any disadvantage.

Joe Sadoski Over a year ago

I'm not a big fan of using safe=false, but in my application, I was able to put the list into a dictionary to make the JsonResponse serializer happy: return JsonResponse({ "someObjects": list(SomeModel.objects.all().values()) })

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Method-1: Use Django's `python` serializer

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