Django-Rest-Framework AssertionError HTTPresponse预期

Question

当我使用curl在终端上执行以下命令时

curl -X POST http://myuser:mypassword@myweb.com:8000/call/make-call/ -d "tutor=1&billed=1"

我收到以下错误

AssertionError at/call/make-call /期望a Response, HttpResponse或者HttpStreamingResponse从视图返回,但是收到了<type 'NoneType'>

我的views.py是

@api_view(['GET', 'POST'])
def startCall(request):

    if request.method == 'POST':

        serializer = startCallSerializer(data=request.DATA)

        if serializer.is_valid():

            serializer.save()

            return Response(serializer.data, status=status.HTTP_201_CREATED)

        else:

            return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

我的serializer.py是

class startCallSerializer(serializers.ModelSerializer):

    class Meta:
        model = call
        fields = ('tutor', 'billed', 'rate', 'opentok_sessionid')

我的urls.py是

urlpatterns = patterns(
    'api.views',
    url(r'^call/make-call/$','startCall', name='startCall'),
)

Answer 1

如果request.method == 'POST'测试失败,该函数不会返回响应.(这是在GET请求上)

@api_view(['GET', 'POST'])
def startCall(request):

    if request.method == 'POST':
        serializer = startCallSerializer(data=request.DATA)

        if serializer.is_valid():
            serializer.save()
            return Response(serializer.data, status=status.HTTP_201_CREATED)
        else:
             return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)
    #Return this if request method is not POST
    return Response({'key': 'value'}, status=status.HTTP_200_OK)