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

/ call / make-call /的AssertionError预计Response， {}返回HttpResponse或HttpStreamingResponse 查看，但收到了<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GET请求）

@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)

Answer 2

添加

#Return this if request method is not POST
    return Response(json.dumps({'key': 'value'},default=json_util.default))

如果您没有在应用程序开发中构建错误代码。

我的完整代码：

@csrf_exempt
@api_view(['GET','POST'])
def uploadFiletotheYoutubeVideo(request):
    if request.method == 'POST': 
        file_obj = request.FILES['file']#this is how Django accepts the files uploaded. 
        print('The name of the file received is ')
        print(file_obj.name)
        posteddata = request.data
        print("the posted data is ")
        print(posteddata)
        response = {"uploadFiletotheYoutubeVideo" : "uploadFiletotheYoutubeVideo"}
        return Response(json.dumps(response, default=json_util.default))
    #Return this if request method is not POST
    return Response(json.dumps({'key': 'value'},default=json_util.default))

Answer 3

编辑如下所示的视图应该可以

@api_view(['GET', 'POST'])
def startCall(request):
    if request.method == 'POST':
    serializer = startCallSerializer(data=request.data)
    data={}
    if serializer.is_valid():
        datas = serializer.save()
        data['tutor']=datas.tutor
        data['billed']=datas.billed
        data['rate']=datas.rate


    else:
        return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)
    return Response(data)

Django-Rest-Framework AssertionError HTTPresponse预期

3 个答案: