How to serialize to JSON a list of model objects in django/python

Question

I am trying to serialize a list of model object defined as:

class AnalysisInput(models.Model):
    input_user = models.CharField(max_length=45)
    input_title = models.CharField(max_length=45)
    input_date = models.DateTimeField()
    input_link = models.CharField(max_length=100)

I wrote a custom serializer (encoder) for json.dumps():

class AnalysisInputEncoder(json.JSONEncoder):
    def default(self, obj):
        if isinstance(obj, AnalysisInput):
            return { "input_id" : obj.id,
                    "input_user" : obj.input_user,
                    "input_title" : obj.input_title,
                    "input_date" : obj.input_date.isoformat(),
                    "input_link" : obj.input_link }
        return json.JSONEncoder.default(self, obj)

When I serialize only one object, I am able to do it. When I try to serialize a list of object I get

[ objects..] is not JSON serializable

I searched but I didn't find where to work on.. I was thinking about writing a custom serializer also for list of model object.

Martijn Pieters · Accepted Answer · 2015-11-17 08:12:31Z

50

A custom encoder is not called recursively. You are actually better off not using a custom encoder, and instead convert your objects to simple python types before serializing.

You could add a as_json or similarly named method to your model and calling that every time you need a JSON result:

class AnalysisInput(models.Model):
    input_user = models.CharField(max_length=45)
    input_title = models.CharField(max_length=45)
    input_date = models.DateTimeField()
    input_link = models.CharField(max_length=100)

    def as_json(self):
        return dict(
            input_id=self.id, input_user=self.input_user,
            input_title=self.input_title, 
            input_date=self.input_date.isoformat(),
            input_link=self.input_link)

Then in your view:

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

# a list of results
results = [ob.as_json() for ob in resultset]
return HttpResponse(json.dumps(results), content_type="application/json")

edited Nov 17, 2015 at 8:12

answered Oct 23, 2012 at 13:11

Martijn Pieters

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7 Comments

gc5 Over a year ago

It works, flawlessly :).. may I ask to you why I should not use custom encoders?

Martijn Pieters Over a year ago

@Francesco: you'd have to write something that fully recurses over all lists, tuples and mappings to find your model objects. In your views, however, you know exactly where your model objects are located already, leading to easier to maintain code. You could combine the two techniques, of course; pyramid's JSON renderer supports __json__ methods on results for example.

Ghasem Over a year ago

When I try it, I get this exception > unbound method as_json() must be called with Category instance as first argument

Martijn Pieters Over a year ago

@AlexJolig: you call it on actual instances of your model, not the model itself.

Ahsan Over a year ago

Looks great! Should the method name/signature be as_dict(self) instead since it's not really returning a json...

|

Envy · Accepted Answer · 2021-06-22 17:08:58Z

22

The best way I found to serialize your Django models is by using django.core.serializers to serialize your model list into JSON, XML, or YAML. No custom serialization code required! Documentation is here: https://docs.djangoproject.com/en/dev/topics/serialization/

Here is my implementation:

lead/models.py:

from django.db import models

class Lead(models.Model):
    name = models.CharField(max_length=50)
    email = models.CharField(max_length=256)
    phone = models.CharField(max_length=20)
    twitter_handle = models.CharField(max_length=20)
    github_handle = models.CharField(max_length=20)

lead/views.py:

from django.http import HttpResponse
from django.core import serializers
from lead.models import Lead

def index(request):
    leads_as_json = serializers.serialize('json', Lead.objects.all())
    return HttpResponse(leads_as_json, content_type='application/json')

The end result:

[{"pk": 1, "model": "lead.lead", "fields": {"twitter_handle": "johndoe", "name": "John Doe", "phone": "1(234)567-8910", "email": "[email protected]", "github_handle": "johndoe"}}]

edited Jun 22, 2021 at 17:08

Envy

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answered May 17, 2013 at 19:19

matthewfisher

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1 Comment

Kalidass Over a year ago

how to customise the response ?

Cody · Accepted Answer · 2017-11-01 11:08:44Z

2

Simplest solution:

def index(request):
    data = serializers.serialize('json', Product.objects.all())
    return HttpResponse(data, content_type='application/json')

answered Nov 1, 2017 at 11:08

Cody

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