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社区首页 >问答首页 >Django RESTful API错误“类型对象‘用户’没有属性'_meta'”

Django RESTful API错误“类型对象‘用户’没有属性'_meta'”
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Stack Overflow用户
提问于 2015-07-07 00:38:35
回答 2查看 1.6K关注 0票数 3

我正在尝试用django-rest框架和RESTful构建Django-rest框架 API,然后一步一步地在Django-rest框架上编写正式的django-rest框架教程,我遇到了这个错误

代码语言:javascript
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Traceback:
File "/usr/local/lib/python2.7/dist-packages/django/core/handlers/base.py" in get_response
132. response = wrapped_callback(request,  *callback_args, **callback_kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/views/decorators/csrf.py" in wrapped_view
58.         return view_func(*args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/views/generic/base.py" in view
  71.             return self.dispatch(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/rest_framework/views.py" in   dispatch
456.             response = self.handle_exception(exc)
File "/usr/local/lib/python2.7/dist-packages/rest_framework/views.py" in   dispatch
444.             self.initial(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/rest_framework/views.py" in    initial
359.         self.check_permissions(request)
File "/usr/local/lib/python2.7/dist-packages/rest_framework/views.py" in check_permissions
304.             if not permission.has_permission(request, self):
File "/usr/local/lib/python2.7/dist-packages/rest_framework/permissions.py" in has_permission
125.         perms = self.get_required_permissions(request.method, queryset.model)
File "/usr/local/lib/python2.7/dist-packages/rest_framework/permissions.py" in get_required_permissions
104.             'app_label': model_cls._meta.app_label,

Exception Type: AttributeError at /users/
Exception Value: type object 'User' has no attribute '_meta'

这是我的密码

urls.py

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from django.conf.urls import  url
from rest_framework.urlpatterns import format_suffix_patterns
from tutorial import views

urlpatterns = [
    # Examples:

    url(r'^users/$', views.UsersList.as_view()),
    url(r'^users/(?P<pk>[0-9]+)/$', views.UserDetail.as_view()),
]


urlpatterns = format_suffix_patterns(urlpatterns)

models.py

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import uuid
from cassandra.cqlengine import columns
from cassandra.cqlengine.models import Model
# Create your models here.

class User(Model):
    id = columns.UUID(primary_key=True, default=uuid.uuid4)
    username = columns.Text(max_length=30, required=True)
    password = columns.Text(max_length=55, required=True)
    gender = columns.UUID(default=0)
    languages = columns.List(value_type=columns.UUID)
    friends = columns.Set(value_type=columns.UUID)

serializers.py

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from rest_framework import serializers
from tutorial.models import User

class UserSerializer(serializers.ModelSerializer):
    class Meta:
        model = User
        fields = ('username', 'position', 'gender', 'status', 'language')

views.py

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from tutorial.models import User
from tutorial.serializers import UserSerializer
from rest_framework import generics

class UsersList(generics.ListCreateAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

class UserDetail(generics.RetrieveUpdateDestroyAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

提前感谢您的支持。

EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2015-07-07 00:58:29

据我所知,cassandra的模型没有实现元类,这是使用像rest_framework这样的应用程序所必需的。您需要实现django兼容的模型来使用rest_framework。

正如django-cassandra的开发人员所指出的:

cqlengine即将被合并到python中。在此之后,我们可以考虑一下,但重要的是要知道, django -cassandra引擎从来没有打算与django模型完全兼容。

Github发行这里

票数 1
EN

Stack Overflow用户

发布于 2015-10-30 09:34:15

请参阅此项目https://github.com/jbatalle/django-cassandra-engine/tree/master/testproject

它使用serializers.Serializer而不是serializers.ModelSerializer。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/31257955

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