/api/register/ 'module' 对象的类型错误不可调用

Question

我正试图使用django-rest框架注册用户，但这是我遇到的错误，请帮助识别问题

/api/register处的类型错误/"module"对象不可调用请求方式：POST请求URL:http://127.0.0.1:8000/api/register/Django版本：3.1.5异常类型：TypeError异常值："模块"对象不可调用异常位置：C:\Users\ben\PycharmProjects\buddyroo\lib\site packages\rest_framework\generics.py，第110行，在get_serializer中Python可执行文件：C:\Users\ben\PycharmProjects\buddyroo\Scripts\Python.exePython版本：3.8.5

下面是RegisterSerializer

from django.contrib.auth.password_validation import validate_password
from rest_framework import serializers
from django.contrib.auth.models import User
from rest_framework.validators import UniqueValidator

class RegisterSerializer(serializers.ModelSerializer):
email = serializers.EmailField(
required=True,
validators=[UniqueValidator(queryset=User.objects.all())]
)
password = serializers.CharField(write_only=True, required=True, validators=[validate_password])
password2 = serializers.CharField(write_only=True, required=True)
class Meta:
model = User
fields = ('username', 'password', 'password2', 'email', 'first_name', 'last_name')
extra_kwargs = {
'first_name': {'required': True},
'last_name': {'required': True}
}
def validate(self, attrs):
if attrs['password'] != attrs['password2']:
raise serializers.ValidationError({"password": "Password fields didn't match."})
return attrs
def create(self, validated_data):
user = User.objects.create(
username=validated_data['username'],
email=validated_data['email'],
first_name=validated_data['first_name'],
last_name=validated_data['last_name']
)
user.set_password(validated_data['password'])
user.save()
return user

和RegisterView.py

from django.contrib.auth.models import User
from rest_framework import generics
from rest_framework.permissions import IsAuthenticated, AllowAny  # <-- Here
from rest_framework.response import Response
from rest_framework.views import APIView
from api import UsersSerializer, RegisterSerializer

class RegisterView(generics.CreateAPIView):
queryset = User.objects.all()
serializer_class = RegisterSerializer
permission_classes = (AllowAny,)

Answer 1

我假设定义RegisterSerializer的模块(文件(的名称是RegisterSerializer.py.

如果是这种情况，则在RegisterView.py中导入的是模块RegisterSerializer，而不是类。

所以，它应该是

from api.RegisterSerializer import RegisterSerializer

在Python中，一个模块中通常有多个类，因此我建议您将模块重命名为serializers.py和views.py，并将所有的序列化程序和视图放在那里。

当然，如果它们很多，您可以将其拆分并创建序列化程序/视图包，并在其中放置几个序列化程序/查看模块：user_serializers.py、whaterver_serializer.py…