I want to write a rest method without model so that I can send a csv file using python requests module. This csv file should be remotely accessed from the server.
For example - I have logged in to my project using requests and get the cookies and headers so that I can pass it to the following requests method..
files = {'file': open('test.csv', 'rb')}
response = requests.post(url, files=files, headers=api_headers,
cookies=api_cookies)
So this url should be : call for that rest method.
views.py file :
class FileUploadView(APIView):
parser_classes = (FileUploadParser,)
def post(self, request, format=None):
csvfile = request.data['file']
#reader = csv.DictReader(csvfile)
#for r in reader:
#print(r)
return Response(status=204)
Just to note - I am sending a csv file using requests module.
Can anyone please help me on how to write this rest method?
Normal django view
def myview(request):
f = request.FILES['file']
with open('some/folder/name.txt', 'wb+') as destination:
#f.name or f.filename (dont know which one)will get filename.So you can replace it name.txt
for chunk in f.chunks():
destination.write(chunk)
return JsonResponse({"message": "Uploaded!"})
UPDATE
# views.py
class FileUploadView(views.APIView):
parser_classes = (FileUploadParser,)
def post(self, request, filename, format=None):
file_obj = request.data['file']
# ...
# do some stuff with uploaded file
# ...
return Response(status=200)
# urls.py
urlpatterns = [
# ...
url(r'^upload/(?P<filename>[^/]+)$', FileUploadView.as_view())
]
Then
url = 'http://127.0.0.1:8000/upload/test.csv' #filename should be in url
files = {'file': open('test.csv', 'rb')}
response = requests.post(url, files=files, headers=api_headers,
cookies=api_cookies)
This would do the trick for you. http://www.django-rest-framework.org/api-guide/parsers/#fileuploadparser
# views.py
class FileUploadView(views.APIView):
parser_classes = (FileUploadParser,)
def post(self, request, filename, format=None):
file_obj = request.data['file']
# ...
# do some stuff with uploaded file
# ...
return Response(status=204)
# urls.py
urlpatterns = [
# ...
url(r'^upload/(?P<filename>[^/]+)$', FileUploadView.as_view())
]
# test with this curl
curl -X POST -S -H -F "file=@something.jpg;type=image/jpg" 127.0.0.1:8000/upload/myfile/