从视图（APIView）中类的url读取字符串

Question

我有一个像我的模式如下网址

urlpatterns = [
...
    path('rest-auth/registration/account-confirm-email/<str:key>', VerifyEmailView.as_view(), name='account_confirm_email'),
...
]

我需要能够传递钥匙（“美育：1ggiiI：923vm1cLboPQk9TX1f0aywDWBs8“）来自此示例url：

http://127.0.0.1:8000/rest-auth/registration/account-confirm-email/MjU:1ggiiI:923vm1cLboPQk9TX1f0aywDWBs8/

以下视图：

class VerifyEmailView(APIView):
    permission_classes = (AllowAny,)
    allowed_methods = ('POST','GET', 'OPTIONS', 'HEAD')
....
    def get(self, request, *args, **kwargs):
        serializer = self.get_serializer(data=request.data)
        serializer.is_valid(raise_exception=True)
        '''key is used here'''
        self.kwargs['key'] = serializer.validated_data['key']
        confirmation = self.get_object()
        confirmation.confirm(self.request)
....

有人能告诉我如何在这个视图中访问我的密钥吗？你知道吗

Answer 1

当您使用url参数时，您可以在get()中直接将其作为函数参数获取，请参见下面的示例：

class VerifyEmailView(APIView):
    permission_classes = (AllowAny,)
    allowed_methods = ('POST','GET', 'OPTIONS', 'HEAD')
....
    def get(self, request, key):
        serializer = self.get_serializer(data=request.data)
        serializer.is_valid(raise_exception=True)
        '''key is used here'''
        self.key = key
        confirmation = self.get_object()
        confirmation.confirm(self.request)
....

在这里你可以直接得到“美育：1ggiiI：923vm1cLboPQk9TX1f0aywDWBs8“inkey变量。你知道吗

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