[Fixed]-Django saving json value to database/model


If I understand your question clearly then
Your view should be something like.

def add_comments(request):
    if 'application/x-www-form-urlencoded' in request.META['CONTENT_TYPE']:
        print 'hi'
        data = json.loads(request.body)
        comment = data.get('comment')
        id = data.get('id')
        title = data.get('title') 
        post = Post.objects.get(id = id)
        com = Comment()
        com. comments = comment
        com.title = post


If you’re using Postgres, you can store json with JSONField (read more), but if not, you need parse json to string and save with CharField/TextField using json.dumps(data). To recovery data, use json string to dict with json.loads(json_string)

Remember to import json lib: import json

Update 2023

Based on @Jarad comment,

JSONField is supported on MariaDB, MySQL, Oracle, PostgreSQL, and SQLite (with the JSON1 extension enabled).




Assuming a model of:

class User(models.Model):
  name = models.CharField()
  phone_number = models.CharField()

Sending json of {“name”:”Test User”, “phone_number”:”123-456-7890″}

In the view you could do the following to save it to the database.

def SaveUser(request):
  body_unicode = request.body.decode('utf-8')
  body = json.loads(body_unicode)
  u = User(**body)
  return JsonResponse({"result": "OK"})


If you want to store the intact JSON, try using the django-jsonfield project: https://github.com/dmkoch/django-jsonfield


according to Django doc you can use :

from django.contrib.postgres.fields import JSONField
from django.db import models

class Dog(models.Model):
    name = models.CharField(max_length=200)
    data = JSONField()

    def __str__(self):
        return self.name

then create with this :

Dog.objects.create(name='Rufus', data={
     'breed': 'labrador',
     'owner': {
         'name': 'Bob',
         'other_pets': [{
             'name': 'Fishy',

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