需要在django API中存储每个user_id更多的车辆

我已经通过使用抽象用户定制了内置用户模型。我制作了另一款与用户模型具有一对一关系的模型车辆，类似于每个用户的车辆。但它存储每个user_id的单个车辆。我需要它为每个用户存储多辆车。我的代码是下面给出 -需要在django API中存储每个user_id更多的车辆

Models.py

class User(AbstractUser): 
    phone_number = models.IntegerField(null=True) 
    otp = models.IntegerField(null=True) 


class Vehicle(models.Model): 
    user = models.OneToOneField(User) 
    maker = models.CharField(max_length=500, blank=True) 
    car_model = models.CharField(max_length=30, blank=True)

Serializer.py

class VehicleSerializer(serializers.ModelSerializer): 
user_id = serializers.IntegerField() 
class Meta: 
    model = Vehicle 
    fields = ('maker', 'car_model', 'user_id')                  

def create(self, validated_data): 
    maker = validated_data.pop('make', None) 
    car_model = validated_data.get('car_model', None) 
    user_id = validated_data.get('user_id', None) 
    instance = self.Meta.model(**validated_data) 

    if user_id is not None: 
     instance.user_id = user_id 
     instance.save() 
    if car_model is not None: 
     instance.car_model = car_model 
     instance.save() 
    return instance

Views.py

@api_view(['POST']) 
def vehicle_add(request): 
car_model = request.data.get('car_model') 
user_id = request.user.id 
serializer = VehicleSerializer(data={'user_id': user_id, 'car_model': car_model}) 
if serializer.is_valid(): 
    serializer.save() 
    message = "i suceeded" 
    return Response(message, status=status.HTTP_201_CREATED) 
else: 
    return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

我只需要知道如何存储多部车辆在单个user_id下。谢谢

来源

2017-08-16 Rishabh Pandey