[Django]-FileUploadParser doesn't get the file name

by

37👍

✅

In django REST framework. we have components like Parsers, Renderers and Serializers.

  • The responsibility of Parsers is to parse the data that is sent by request methods GET, POST and PUT, etc.

  • Default parser used in django REST is ‘JSONParser‘. It only parses the data JSON data[numbers, string, date]. It ignores the data like FILES.

  • In order to parse the FILES we need to use parsers like “MultiPartParser” or “FormParser“.

    Example Code :

        from rest_framework.parsers import MultiPartParser
    from rest_framework.response import Response
    from rest_framework.views import APIView

    class ExampleView(APIView):
        """
        A view that can accept POST requests with JSON content.
        """
        parser_classes = (MultiPartParser,)

        def post(self, request, format=None):
            # to access files
            print request.FILES
            # to access data
            print request.data
            return Response({'received data': request.data})

When we use property request.data then parser will parse the data.

References: Django REST Docs, Django REST Github

👤anjaneyulubatta505

4👍

I face the same problem.The problem error message shows:

{“detail”:”Missing filename. Request should include a
Content-Disposition header with a filename parameter.”}
I do all the steps above my answer but it doesn’t work.Finally,

I find the reason is in the backend of viewset.

it shows like this

parser_classes = (FileUploadParser, MultiPartParser, FormParser)

and I remove the FileUploadParser

  parser_classes = ( MultiPartParser, FormParser)

and It works, so I think you should pay more attention to it

👤Deft-pawN

3👍

Adding

Content-Disposition: attachment; filename=<insert-your-file-name>

to the header has solved my issue, and then in the view method where you access this file in the following manner:

file = self.request.FILES['file']
👤Ali H. Kudeir

2👍

On the second error Missing filename. Request should include a Content-Disposition header with a filename parameter. using postman I removed the header Content-Type :multipart/form-data and was successful.

It seems the issue is the custom Content-Type header overrides the default Content-Type header that should be sent. Check this thread for reference Content-Type for multipart posts

👤StackEdd

2👍

You don’t need to use MultipartParser or FormParser.

What you need is a serializer with a FileField() like so:

serializers.py:

class FileUploadSerializer(serializers.Serializer):
    # I set use_url to False so I don't need to pass file 
    # through the url itself - defaults to True if you need it
    file = serializers.FileField(use_url=False)

So then when you try to access file here below, you’ll have a dict with a key named file. Personally, I’d probably name it something more descriptive than just “file”, but that’s up to you.

from .serializers import FileUploadSerializer

    class FileUploadView(APIView):

       def post(self, request):
           # set 'data' so that you can use 'is_vaid()' and raise exception
           # if the file fails validation
           serializer = FileUploadSerializer(data=request.data)
           serializer.is_valid(raise_exception=True)
           # once validated, grab the file from the request itself
           file = request.FILES['file']
👤Paul Jurczyk

0👍

in your views.py change the parser like this

parser_classes = (JSONParser, MultiPartParser)

👤Darwin

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