python - 在创建方法 django rest 框架中引发错误

Question

我扔ValidationError在验证我的 clean 中的数据时models.py 中的方法.如何在我的自定义 create 中捕获此错误方法，以便它抛出一个包含此错误详细信息的json对象
道路

{
    "detail":"input is not valid"
}

#models.py
class 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.py
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 ex
    return obj

Answer 1

检查您是否已抛出 serializers.ValidationError不是 ValidationError的 django.core.exceptions .您可以更改您的 create这种方式的方法:

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