Return user_id 作为 post api 的回应

Question

我想 return user_id 使用 post api 创建的每个对象。

我想要

形式的输出

[
{"info": {
        "status": "SUCCESS",
        "message": "User Details Successfully Uploaded"
        },
        "user_id": 10001
    }
]

这是我的代码

Serializer.py

class UserDetailsSerializer(serializers.ModelSerializer):
    """docstring for UserDetailsSerializer"""
    class Meta:
        model = UserDetails
        fields = ['user_id', 'user_email', 'user_full_name', 'user_token', 'patient_id', 'user_preferences']
        read_only_fields = ['user_id']

Views.py

class UserDetailsViewSet(APIView):

    def get(self, request):
        queryset= UserDetails.objects.all()
        serializer_class=UserDetailsSerializer(queryset,many=True)
        return Response(serializer_class.data)

    def post(self, request):
        serializer_class_post=UserDetailsSerializer(data=request.data)
        if serializer_class_post.is_valid():
            try:
                serializer_class_post.save()
                return Response([{"info": {
                    "status": "SUCCESS",
                    "message": "User Details Successfully Uploaded"
                },
                    "user_id": serializer_class_post.data
                }], status=status.HTTP_201_CREATED)
            except IntegrityError as e:
                return Response([{"info": {
                    "status": "Error",
                    "message": "Error Uploading User Details"
                }
                }], status=status.HTTP_400_BAD_REQUEST)
        return Response(serializer_class_post.errors, status=status.HTTP_400_BAD_REQUEST)`

models.py

class UserDetails(models.Model):
    user_id = models.IntegerField(primary_key= True)
    user_email = models.CharField(max_length=254)
    user_full_name = models.CharField(max_length=200)
    user_token = models.CharField(max_length=200)
    patient_id = models.CharField(max_length=12, unique=True)
    user_preferences = models.CharField(max_length=200)

    class Meta:
        managed = False
        db_table = 'walnut_users'

    def __str__(self):
        return self.user_id

Answer 1

将您的 post 更改为以下内容：

def post(self, request):
    serializer_class_post=UserDetailsSerializer(data=request.data)
    if serializer_class_post.is_valid():
        try:
            obj = serializer_class_post.save()
            return Response([{"info": {
                "status": "SUCCESS",
                "message": "User Details Successfully Uploaded"
            },
                "user_id": obj.pk
            }], status=status.HTTP_201_CREATED)
        except IntegrityError as e:
            return Response([{"info": {
                "status": "Error",
                "message": "Error Uploading User Details"
            }
            }], status=status.HTTP_400_BAD_REQUEST)
    return Response(serializer_class_post.errors, status=status.HTTP_400_BAD_REQUEST)`

serializer.save() returns 新创建的对象；只需使用它并在您的回复中获得 pk。

此外，API 最佳做法是 return 完整的对象表示；不仅仅是身份证。理想情况下，我会做你所做的：{"data": serializer_post_data.data} -> 这将 return 完整的嵌套表示。您应该仍然能够以 data.user_id 访问 user_id。但是，如果您确定您的用例，上面的代码将起作用。

因为 user_id 是一个 IntegerField 和主键，所以 Django 希望您提供该字段的值。您在哪里为 user_id 赋值 - 您的 API 是否也 post 为 user_id 赋值。如果要自动分配值，请将其更改为 user_id = models.AutoField(primary_key=True)

Return user_id 作为 post api 的回应

Return user_id as response of post api

django

django-models

django-views

django-serializer

django-rest-framework