Python Django Rest框架关系&;超链接API问题
我正在尝试django rest框架。在我进入教程的关系和超链接API部分之前,一切都很顺利。我现在遇到的一个错误是:Python Django Rest框架关系&;超链接API问题,python,django,django-rest-framework,Python,Django,Django Rest Framework,我正在尝试django rest框架。在我进入教程的关系和超链接API部分之前,一切都很顺利。我现在遇到的一个错误是: 在/api/users/“^\(?p[a-z0-9]+)\(?p[a-z0-9]+)$”处配置不当不是有效的正则表达式:将组名u'format'重新定义为group2;was第1组 我试着做了一些研究,但似乎找不到任何东西,我越是搞砸了,出错的就越多 这是我的密码: 模块.py class Home(models.Model): user = models.Foreig
在/api/users/“^\(?p[a-z0-9]+)\(?p[a-z0-9]+)$”处配置不当不是有效的正则表达式:将组名u'format'重新定义为group2;was第1组
我试着做了一些研究,但似乎找不到任何东西,我越是搞砸了,出错的就越多
这是我的密码:
模块.py
class Home(models.Model):
user = models.ForeignKey(User)
#address ect
序列化程序.py
class UserSerializer(serializers.HyperlinkedModelSerializer):
username = serializers.HyperlinkedRelatedField(many=True, read_only=True, view_name='home-detail')
class Meta:
model = User
fields = ('url', 'username', 'home')
class HomeSerializer(serializers.HyperlinkedModelSerializer):
owner = serializers.Field(source='owner.username')
highlight = serializers.HyperlinkedIdentityField(view_name='home-highlight', read_only=True, format='html')
class Meta:
model = Home
fields = ('url', 'owner', 'postcode')
api.py
@api_view(('GET',))
def api_root(request, format=None):
return Response({
'users': reverse('user-list', request=request, format=format),
'homes': reverse('home-list', request=request, format=format)
})
class UserList(generics.ListCreateAPIView):
queryset = User.objects.all()
serializer_class = UserSerializer
class UserDetail(generics.RetrieveUpdateDestroyAPIView):
queryset = User.objects.all()
serializer_class = UserSerializer
class HomeList(generics.ListCreateAPIView):
queryset = Home.objects.all()
serializer_class = HomeSerializer
class HomeDetail(generics.RetrieveUpdateDestroyAPIView):
queryset = Home.objects.all()
serializer_class = HomeSerializer
class HomeHighlight(generics.GenericAPIView):
queryset = Home.objects.all()
renderer_classes = (renderers.StaticHTMLRenderer,)
def get(self, request, *args, **kwargs):
snippet = self.get_object()
return Response(snippet.highlighted)
url.py
urlpatterns = format_suffix_patterns([
url(r'^$', api.api_root),
url(r'^users/$',
api.UserList.as_view(),
name='user-list'),
url(r'^users/(?P<pk>[0-9]+)/$',
api.UserDetail.as_view(),
name='user-detail'),
url(r'^home/$',
api.HomeList.as_view(),
name='home-list'),
url(r'^home/(?P<pk>[0-9]+)/$',
api.HomeDetail.as_view(),
name='home-detail'),
url(r'^home/(?P<pk>[0-9]+)/highlight/$',
api.HomeHighlight.as_view(),
name='home-highlight')
])
urlpatterns += [
url(r'^api-auth/', include('rest_framework.urls',
namespace='rest_framework')),
]
urlpatterns = format_suffix_patterns(urlpatterns)
urlpatterns=格式\后缀\模式([
url(r'^$',api.api_根),
url(r“^users/$”,
api.UserList.as_view(),
name='user-list'),
url(r'^users/(?P[0-9]+)/$”,
api.UserDetail.as_view(),
name='user-detail'),
url(r“^home/$”,
api.HomeList.as_view(),
name='home-list'),
url(r'^home/(?P[0-9]+)/$”,
api.HomeDetail.as_view(),
name='home-detail'),
url(r'^home/(?P[0-9]+)/highlight/$”,
api.HomeHighlight.as_view(),
name='home-highlight')
])
URL模式+=[
url(r“^api auth/”,包括('rest\u framework.url',
namespace='rest_framework'),
]
urlpatterns=格式\后缀\模式(urlpatterns)
您调用了两次format\u suffix\u patterns
,因此Django不知道如何解析URL,因为有两个format
组
您不应该需要第一个调用,因为第二个调用会为您处理它(并允许
令牌身份验证
保留后缀)。至少将router.DefaultRouter()更改为router.SimpleRouter()在URL.py文件中,我得到了相同的错误,因为DefaultRouter
在其实现中调用了format\u suffix\u patterns
。@Seth可以通过添加router.include\u format\u suffix=False
,在router=routers.DefaultRouter()
之后解决。我有URLPterns=format\u suffix\u patterns(URLPterns)这是在url.py中导致的