Question

我目前正在使用django-rest-framewok版本处理Django项目。我面临一个奇怪的问题，说意外的关键字参数命名版本。

views.py

class UserViewSet(viewsets.ModelViewSet):

    queryset = ScreenShotUser.objects.all()
    serializer_class = UserSerializer


def getMe(self,request):
      user = ScreenShotUser.objects.get(pk=request.user.id)
      return Response(user)

项目url.py

router = routers.DefaultRouter()
router.register(r 'users', views.UserViewSet)
urlpatterns = [
  url(r '^api/(?P<version>(v1|v2))/accounts/', include('Accounts.urls')),
]

和帐户应用程序的urls.py

 urlpatterns = [
    url(r'^', include(router.urls)),
    url(r'^me/$', views.getMe),
 ]

当我致电http://127.0.0.1:8000/api/v1/accounts/me/时我得到了这样的错误

TypeError at /api/v1/accounts/me/
getMe() got an unexpected keyword argument 'version'
Request Method: GET
Request URL:    http://127.0.0.1:8000/api/v1/accounts/me/
Django Version: 1.9.4
Exception Type: TypeError
Exception Value:    
getMe() got an unexpected keyword argument 'version'
Exception Location: /Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/django/core/handlers/base.py in get_response, line 147
Python Executable:  /Library/Frameworks/Python.framework/Versions/3.5/bin/python3.5
Python Version: 3.5.1
Python Path:    
['/Working_repo/screenshot/ScreenShot',
 '/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/django_contrib_requestprovider-1.0.1-py3.5.egg',
 '/Working_repo/screenshot/ScreenShot',
 '/Library/Frameworks/Python.framework/Versions/3.5/lib/python35.zip',
 '/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5',
 '/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/plat-darwin',
 '/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/lib-dynload',
 '/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages']
Server time:    Mon, 28 Mar 2016 13:12:54 +0000

Django专家请给我一些我遗漏的提示。 TIA

修改我有解决方案概念形式尚旺的答案，这将是我的解决方案

@api_view(['GET'])
def getMe(request, version):
    user = ScreenShotUser.objects.get(pk=request.user.id)
    result = UserSerializer(user).data
    return Response(result)

Answer 1

如果您在网址中定义了(?P<version>(v1|v2))，则表示您的views.py方法需要参数version。但是您的getMe方法没有将其作为参数，因此错误。轻松修复将删除网址定义中的命名参数。

修改

你应该看看django doc about named group，在python中(?P<name>pattern)表示一个名为group的正则表达式。如果你在django url定义中有这个，那就意味着你将一个参数传递给views.py方法。我不认为你的views.py方法正在使用version参数，所以你不应该在url中定义。你这样做：

url(r '^api/v1|v2/accounts/', include('Accounts.urls')),

如果您无法删除version参数，因为其他一些方法确实需要它，只需在getMe方法中定义它并且不要将它用于任何事情：

def getMe(self, request, version): user = ScreenShotUser.objects.get(pk=request.user.id) return Response(user)

在how url dispatcher is working上阅读django doc中的更多详细信息。

/ api / v1 / accounts / me / getMe（）中的TypeError获得了一个意外的关键字参数＆＃39;版本＆＃39;

1 个答案: