Django REST Framework ValidationError always returns 400

问题

I am trying to force ValidationError to return a different status code than 400. This is what I did:

class MyValidationError(ValidationError):
    status_code = HTTP_403_FORBIDDEN

and then in a serializer:

def validate_field(self, value):
    raise MyValidationError

Why do I get 400 here instead of 403? An interesting thing is that if I use PermissionDenied with a custom status code (I tried 204) instead of ValidationError, it works as expected.

回答1:

The Django RestFramework serializer validates all possible fields and finally returns set of errors.

That is, suppose you are expecting two validation errors in serializer, in that one validation error is raised by MyValidationError. In that case DRF obviously return HTTP 400 status code, because the design patterns of DRF not raising individual validation errors.

The serializer validation process done inside the is_valid() method and it raises ValidationError at the end of the method.

def is_valid(self, raise_exception=False):
    ....code
    ....code
    if self._errors and raise_exception:
        raise ValidationError(self.errors)

    return not bool(self._errors)

and the ValidationError class raises HTTP 400

class ValidationError(APIException):
    status_code = status.HTTP_400_BAD_REQUEST
    default_detail = _('Invalid input.')
    default_code = 'invalid'
    .... code

Why PermissionDenaid returns custom status code?

In the is_valid() (source code) method, it catches only ValidationError

if not hasattr(self, '_validated_data'):
    try:
        self._validated_data = self.run_validation(self.initial_data)
    except ValidationError as exc:
        self._validated_data = {}
        self._errors = exc.detail
    else:
        self._errors = {}

At that time, DRF directly raises a PermissionDenaid exception and returns its own status code, which is customized by you

Conclusion

DRF Serializer ValidationError never returns status code other that HTTP 400 because it catches other sub validation error exceptions(if any) and atlast raises a major ValidationError which return HTTP 400 status code by it's design pattern

Reference:
is_valid() source code
ValidationError class source code

回答2:

yo bruh just create your own exception class and reraise it in serializer

class ValidationError422(APIException):
    status_code = status.HTTP_422_UNPROCESSABLE_ENTITY


class BusinessSerializer(serializers.ModelSerializer):
    class Meta:
        model = Business
        fields = ('id', 'name')

    def is_valid(self, raise_exception=False):
        try:
            return super(BusinessSerializer, self).is_valid(raise_exception)
        except ValidationError as e:
            raise ValidationError422(detail=e.detail)

来源：https://stackoverflow.com/questions/51566095/django-rest-framework-validationerror-always-returns-400

标签

django-rest-framework

http-status-codes

serializer