[Fixed]-Django: Create and save a model using JSON data

24👍

You could just do,

PartOne.objects.create(**json)

You can use the **kwargs syntax when calling functions by constructing a dictionary of keyword arguments and passing it to your function.

This is documented on section 4.7.4 of python tutorial., under unpacking argument lists.

Also, note that the same dict is not passed into the function. A new copy is created, so “json” is not kwargs.

3👍

You may also want to take a look at modelform_factory if you want to run more validation on your data or have more control over input. You can also do good stuff like attaching files.

from django.forms.models import modelform_factory

form = modelform_factory(PartOne, fields=('gender', 'gender_na', 'height', 'height_na',))
populated_form = form(data=my_json)

if populated_form.is_valid():
    populated_form.save()
else:
    # do something, maybe with populated_form.errors
👤whp

0👍

There is another way I use, create a method in the model (serialize) and then use JsonResponse to return:

class User(AbstractUser):
    u_username = models.CharField(max_length=100, default=None, blank=True, null=True)
    p_pic = models.CharField(max_length=500, default=None, blank=True, null=True)
    p_comment = models.CharField(max_length=200, default=None, blank=True, null=True)

    def serializeProfile(self):
        return {
            "pk": self.pk,
            "username": self.username,
            "email": self.email
        }

And then call the function, inside of the jsonRespone:

u_profile = User.objects.get(username='username')
JsonResponse(u_profile.serializeProfile())

The above worked for me.

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