如何在Django REST框架中重写viewsets.ModelViewSet？

Question

所以我的模型很简单

class Face(models.Model):
    uid = models.CharField(max_length=510, primary_key=True)
    photo = models.ImageField(upload_to='face_photos')

序列化程序

class FaceSerializer(serializers.HyperlinkedModelSerializer):
    class Meta:
        model = Face
        fields = ['uid', 'photo']
        extra_kwargs = {'uid': {'required': True}, 'photo': {'required': True}}

视图应该是这样的

class FaceViewSet(viewsets.ModelViewSet):
    queryset = Face.objects.all()
    serializer_class = FaceSerializer
    permission_classes = [permissions.AllowAny]

它是有效的。然而：

1. 我不想要列表、更新、删除选项。只发布和获取
2. 我想在post上添加我的逻辑，所以如果uid存在，则更新，否则创建。。。以及其他处理
3. 我想在发布后定制回复

如何在不失去viewsets.ModelViewSet提供的所有优点（如验证、RESTAPI web视图中自动生成的HTML字段等）的情况下实现这一切

Answer 1

这对我很有用：

class FaceViewSet(mixins.CreateModelMixin, mixins.RetrieveModelMixin, GenericViewSet):
    """
    API endpoint for adding and processing new client (by uid) face
    """
    queryset = Face.objects.all()
    serializer_class = FaceSerializer
    permission_classes = [permissions.AllowAny]

    def create(self, request):
        if "uid" in request.POST:
            try:
                instance = Face.objects.get(pk=request.POST['uid'])
                serializer = FaceSerializer(
                    instance=instance,
                    data=request.data
                )
            except Face.DoesNotExist:
                serializer = FaceSerializer(data=request.data)
        else:
            serializer = FaceSerializer(data=request.data)

        serializer.is_valid(raise_exception=True)
        serializer.save()

        return Response(serializer.data)

    def retrieve(self, request, *args, **kwargs):
        instance = self.get_object()
        serializer = FaceSerializer(instance=instance)
        return Response(serializer.data)