Django REST Framework how to specify error code when raising validation error in serializer

Question

I have an API endpoint that allow users to register an account. I would like to return HTTP 409 instead of 400 for a duplicate username.

Here is my serializer:

from django.contrib.auth.models import User
from rest_framework.serializers import ModelSerializer

class UserSerializer(ModelSerializer):
    username = CharField()

    def validate_username(self, value):
        if User.objects.filter(username=value).exists():
            raise NameDuplicationError()
        return value


class NameDuplicationError(APIException):
    status_code = status.HTTP_409_CONFLICT
    default_detail = u'Duplicate Username'

When the error is triggered, the response is: {"detail":"Duplicate Username"}. I realised that if I subclass APIException, the key detail is used instead of username.

I want to have this response instead {"username":"Duplicate Username"}

or I would like to specify a status code when raising a ValidationError:

def validate_username(self, value):
    if User.objects.filter(username=value).exists():
        raise serializers.ValidationError('Duplicate Username', 
                                          status_code=status.HTTP_409_CONFLICT)
    return value

But this does not work as ValidationError only returns 400.

Is there any other way to accomplish this?

Answer 1

You can raise different exceptions like:

from rest_framework.exceptions import APIException
from django.utils.encoding import force_text
from rest_framework import status


class CustomValidation(APIException):
    status_code = status.HTTP_500_INTERNAL_SERVER_ERROR
    default_detail = 'A server error occurred.'

    def __init__(self, detail, field, status_code):
        if status_code is not None:self.status_code = status_code
        if detail is not None:
            self.detail = {field: force_text(detail)}
        else: self.detail = {'detail': force_text(self.default_detail)}

you can use this in your serializer like:

raise CustomValidation('Duplicate Username','username', status_code=status.HTTP_409_CONFLICT)

or

raise CustomValidation('Access denied','username', status_code=status.HTTP_403_FORBIDDEN)

Answer 2

By default, raising serializers.ValidationError will return with HTTP_400_BAD_REQUEST

But sometimes we would like to return ValidationError with normal 200 status code, because some libraries on the client side can't parse json response data while response code is not 200.

I tried this. but it's not worked:

raise serializers.ValidationError({'message':'Invalid  email address'}, code=200)

So we can do this and it works:

res = serializers.ValidationError({'message':'Invalid  email address'})
res.status_code = 200
raise res

Answer 3

Use django-rest-framework custom exception handler http://www.django-rest-framework.org/api-guide/exceptions/

def custom_exception_handler(exc, context=None):
    response = exception_handler(exc, context)
    if response is not None:
         if response.data['detail'] == 'Duplicate Username':
            response.data['username'] = response.data.pop('detail')
        response.status_code = status.HTTP_409_CONFLICT
    return response