Sometimes you need to accept a file from a user, and you don’t want to use django Forms. The details come down to how many files you need to accept and do you know the names of the files.

If the file is from your app (or frontend) and you expect it you’ll know the file name, because you named it. But if it is a user selected file the name will be unknown.

Model

class FileModel(Model):
    file = FileField(upload_to='uploaded_files/' default='')
    created = DateTimeField(auto_now=True)

If the expected file is an image, change it to ImageField.

View

This paragraph shows the base code, for a single file and unknown filename.

class FileUploadView(View):

    def post(self, request):
        file_model = FileModel()
        _, file = request.FILES.popitem()  # get first element of the uploaded files

        file = file[0]  # get the file from MultiValueDict

        file_model.file = file
        file_model.save()

        return HttpResponse(content_type='text/plain', content='File uploaded')

In Python 3.x the request.FILES object is a dictionary that contains a fieldname as a key and a MultiValueDictionary as value. That’s because ex. in your HTML form you can upload multiple files from the same key. This way they will all be under the same key, and their names will be in the MultiValueDict. So the line:

file = file[0]

is only valid when we’re sure that only one file was sent.

Multiple files

In order to accept and save multiple files we need to add a loop to the main logic of our view. The loop will look like this, if the files were sent from the same form field (ex. <input type="file" name="files_to_upload"> ):

_, files = request.FILES.popitem()  # the files have been sent form the same form field (multiple files selected)

for f in file:
    filename = f.name

Alternatively you can use the key name to get the dictionary:

for f in request.FILES.getlist('files_to_upload'):
    filename = f.name

If there are multiple fields in the form, each uploading a file, you need to loop over the items of the request.FILES object.

for field_name, file in request.FILES.iteritems():
    filename = file[0].name