我是Django REST框架的新手。我创建了一个API,它将MP4文件作为输入,对其进行处理并给出响应。我可以通过API浏览器页面上传文件并得到结果。但是,当我试图通过本地python文件调用API时,出现以下错误:
MultiValueDictKeyError at /api/ 'file'
Request Method: POST
API调用:
import requests
import json
url = 'http://182.xx.xx.xx:8000/api/'
video_file = '/home/lueinuser/Downloads/original.mp4' # 30 MB file in my Local system
payload = {'file' : video_file}
r = requests.post(url=url,data=payload)
print(r.text)
视图.py
class ApiCallingListView(generics.ListCreateAPIView):
queryset = ApiCallingModel.objects.all()
serializer_class = ApiCallingSerializer
parser_classes = (MultiPartParser, FormParser)
def post(self, request, format=None, *args, **kwargs):
file_serializer = ApiCallingSerializer(data=request.data)
file = request.FILES['file']
fs = FileSystemStorage()
filename = fs.save(file.name, file)
uploaded_file_url = settings.MEDIA_ROOT + '/' + str(filename)
data_dict = {'video_file': request.FILES}
request.data['response']= api_wrapper(uploaded_file_url)
if file_serializer.is_valid():
file_serializer.save()
return Response(file_serializer.data, status=status.HTTP_201_CREATED)
else:
return Response(file_serializer.errors, status=status.HTTP_400_BAD_REQUEST)
型号.py
class ApiCallingModel(models.Model):
file = models.FileField(blank=False, null=False)
user_id = models.ForeignKey(User, on_delete=models.CASCADE, null=True)
response = models.TextField()
timestamp = models.DateTimeField(auto_now_add=True)
您没有正确发送文件:
您需要
open
这样的文件:并使用
files
命名参数调用request.post
,而不是作为data
:见docs
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