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Python 没有身份验证的Django REST框架_Python_Django_Django Rest Framework - Fatal编程技术网

Python 没有身份验证的Django REST框架

Python 没有身份验证的Django REST框架,python,django,django-rest-framework,Python,Django,Django Rest Framework,我正在尝试使用Django REST框架创建一个基本视图,而不需要身份验证 设置.py THIRD_PARTY_APPS = ( 'south', # Database migration helpers: 'crispy_forms', # Form layouts 'rest_framework', ) REST_FRAMEWORK = { 'DEFAULT_PERMISSION_CLASSES':

我正在尝试使用Django REST框架创建一个基本视图,而不需要身份验证

设置.py

THIRD_PARTY_APPS = (
        'south',  # Database migration helpers:
        'crispy_forms',  # Form layouts
        'rest_framework',
    )

REST_FRAMEWORK = {
                'DEFAULT_PERMISSION_CLASSES': (
                    'rest_framework.permissions.AllowAny',
                )
    }
router = routers.DefaultRouter()
router.register(r'ticket', views.TicketViewSet)

urlpatterns = patterns('',
    url(r'^', include(router.urls)),
    url(r'^test', include('rest_framework.urls', namespace='rest_framework')),

) 
url.py

THIRD_PARTY_APPS = (
        'south',  # Database migration helpers:
        'crispy_forms',  # Form layouts
        'rest_framework',
    )

REST_FRAMEWORK = {
                'DEFAULT_PERMISSION_CLASSES': (
                    'rest_framework.permissions.AllowAny',
                )
    }
router = routers.DefaultRouter()
router.register(r'ticket', views.TicketViewSet)

urlpatterns = patterns('',
    url(r'^', include(router.urls)),
    url(r'^test', include('rest_framework.urls', namespace='rest_framework')),

) 
序列化程序

from django.contrib.auth.models import User, Group
from rest_framework import serializers

class TicketInputSerializer(serializers.HyperlinkedModelSerializer):
    class Meta:
        model = Abc
        fields = ('test')
from django.db import models

class Abc(models.Model):
    test = models.CharField(max_length=12)
型号

from django.contrib.auth.models import User, Group
from rest_framework import serializers

class TicketInputSerializer(serializers.HyperlinkedModelSerializer):
    class Meta:
        model = Abc
        fields = ('test')
from django.db import models

class Abc(models.Model):
    test = models.CharField(max_length=12)
错误

当尝试获取URI时,我得到以下结果:

Page Not Found 404

Using the URLconf defined in config.urls, Django tried these URL patterns, in this order: 
^test ^login/$ [name='login']
^test ^logout/$ [name='logout'] 

有什么想法吗?

你可以访问哪个URL?如果你想访问API端点,你需要转到:
/ticket/
。谢谢,不过我现在有另一个问题,但我会打开一个新的任务