在创建方法 django rest 框架中引发错误

Question

ValidationError我在验证我的clean方法中的数据时抛出models.py. 如何在我的自定义create方法中捕获此错误，以便它以这种方式抛出包含错误详细信息的 json 对象{    "detail":"input is not valid"}#models.pyclass Comment(models.Model):    text = models.CharField(max_length=256)    commenter = models.ForeignKey(User, on_delete=models.SET_NULL)    post = models.ForeignKey(Post, on_delete=models.SET_NULL)    def clean(self, *args, **kwargs):        if containsBadWords(text):            raise ValidationError(_("Be Polite"))#serializer.pydef create(self, validated_data):    request = self.context.get('request', None)    commenter = request.user    try:        obj = Comment.objects.create(            post = validated_data['post'],            commenter = commenter,            text = validated_data['text']        )    except ValidationError as ex:        raise ex    return obj

Answer 1

检查你是否抛出了serializers.ValidationErrornot ValidationErrorof django.core.exceptions。您可以通过以下方式更改您的create方法：

def create(self, validated_data):

    request = self.context.get('request', None)

    commenter = request.user

    try:

        obj = Comment.objects.create(

            post = validated_data['post'],

            commenter = commenter,

            text = validated_data['text']

        )

    except ValidationError as ex:

        raise serializers.ValidationError({"detail": "input is not valid"})

    return obj