How to write Django REST Api without model to send a file using Requests

Question

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?

REST Api doesn't mean it need a Model. You can do normal views to accept the file. — itzMEonTV
– itzMEonTV, Commented Apr 10, 2017 at 9:58
so you want to upload a file using api and you would do some processing or what ever on the file in the View ? — rrmerugu
– rrmerugu, Commented Apr 10, 2017 at 9:58
Yup. it have direct doc django-rest-framework.org/api-guide/parsers/#fileuploadparser or you are not using drf ? — itzMEonTV
– itzMEonTV, Commented Apr 10, 2017 at 10:02

itzMEonTV · Accepted Answer · 2017-04-11 10:59:44Z

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)

itzMEonTV · Accepted Answer · 2017-04-10 11:35:06Z

3

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 "[email protected];type=image/jpg" 127.0.0.1:8000/upload/myfile/

edited Apr 10, 2017 at 11:35

itzMEonTV

20.4k4 gold badges44 silver badges53 bronze badges

answered Apr 10, 2017 at 10:02

rrmerugu

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11 Comments

Sneha Shinde Over a year ago

I have created above function - and when I try to send a file using requests - files = {'file': open('test.pdf', 'rb')}, response = requests.post('127.0.0.1:8000/upload/', files=files, headers=api_headers,cookies=api_cookies). It gives me an 400 error:Bad request.Please check and help.

itzMEonTV Over a year ago

@SnehaShinde Then why did you accept that. Instead of def put, use post

rrmerugu Over a year ago

@Sneha Can u post the log error if any from the console

Sneha Shinde Over a year ago

@itzmeontv Sorry for that,I thought it will work.Anyway I tried replacing put with 'post'.but still it's giving the same error.Am I defining the url correctly - '127.0.0.1:8000/upload/' to the requests?

itzMEonTV Over a year ago

@SnehaShinde try with post(self, request, format=None) if you are not sending filename in url

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