Python Django-Rest-Framework AssertionError HTTPresponse 预期
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Django-Rest-Framework AssertionError HTTPresponse Expected
提问by Eddwin Paz
When i do the following command over Terminal using curl
当我使用 curl 在终端上执行以下命令时
curl -X POST http://myuser:[email protected]:8000/call/make-call/ -d "tutor=1&billed=1"
I get the following error
我收到以下错误
AssertionError at /call/make-call/ Expected a
Response,HttpResponseorHttpStreamingResponseto be returned from the view, but received a<type 'NoneType'>
AssertionError at /call/make-call/ 预期 a
Response，HttpResponse或HttpStreamingResponse从视图返回，但收到一个<type 'NoneType'>
My views.py is
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)
my serializer.py is
我的 serializer.py 是
class startCallSerializer(serializers.ModelSerializer):
class Meta:
model = call
fields = ('tutor', 'billed', 'rate', 'opentok_sessionid')
my urls.py is
我的 urls.py 是
urlpatterns = patterns(
'api.views',
url(r'^call/make-call/$','startCall', name='startCall'),
)
回答by Kwaw Annor
The function does not return a Response if the request.method == 'POST'test fail.
(That is on a GET request)
如果request.method == 'POST'测试失败，该函数不会返回响应。（这是一个 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)
回答by Haris Np
Just add
只需添加
#Return this if request method is not POST
return Response(json.dumps({'key': 'value'},default=json_util.default))
if you don't have an error code built in your application development.
如果您的应用程序开发中没有内置错误代码。
My full code :
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))
回答by Suraj Ghale
Editing the views like below should work
编辑如下所示的视图应该可以工作
@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)
