ValueError: Filename must be a string while uploading file to s3 bucket

Question

I am sending an excel file as a request in postman and need to upload this to s3 . I access the file from request and send it to s3.

@api_view(['POST'])
def excel_upload(request):
    print("request", request)
    excel_file = request.FILES['file'] 
    print("excel_file", excel_file) // this prints the name of the excel file i am sending in request
    upload_to_aws(excel_file,'X1excelsheets','s3_file_name')

and here is the function to upload file to s3.

def upload_to_aws(local_file, bucket, s3_file):
    s3 = boto3.client('s3', aws_access_key_id=ACCESS_KEY,
                      aws_secret_access_key=SECRET_KEY)

    try:
        s3.upload_file(local_file, bucket, s3_file)
        print("Upload Successful")
        return True
    except FileNotFoundError:
        print("The file was not found")
        return False
    except NoCredentialsError:
        print("Credentials not available")
        return False


uploaded = upload_to_aws('local_file', 'bucket_name', 's3_file_name')

I am trying to use this particular post

https://medium.com/bilesanmiahmad/how-to-upload-a-file-to-amazon-s3-in-python-68757a1867c6

to get things done . Error:ValueError: Filename must be a string

Heyo - You are passing file as it is, excel_file = request.FILES['file']. Probably you should save it, and send to aws by "filename". So excel_file='/path/to/file/file.xls' — Dawid Gacek
– Dawid Gacek, Commented Jan 24, 2020 at 6:29
Why so ?I don't want to save it . I want it to be sent directly to aws s3 directly from the request . — horrible_nerd
– horrible_nerd, Commented Jan 24, 2020 at 6:30
Sorry, not - what could be worth trying is: boto3.amazonaws.com/v1/documentation/api/latest/reference/… — Dawid Gacek
– Dawid Gacek, Commented Jan 24, 2020 at 6:44

Trajanov Risto · Accepted Answer · 2020-08-09 21:27:53Z

6

I made my configuration with AWSCLI

S3_BUCKET_NAME = 'YOUR_BUCKET'

s3 = boto3.client('s3')
with open(file, 'rb') as f:
    s3.upload_fileobj(f, S3_BUCKET_NAME, file)

I have been getting the same error so I just used upload_fileobj instead of upload_file. This worked fine for me, you can try it.

answered Aug 9, 2020 at 21:27

Trajanov Risto

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Neelabh Singh · Accepted Answer · 2022-11-14 18:04:06Z

0

Save execel_file locally received from request.POST.get('file_name') in '/tmp' directory or according where your backend code will deploy, because upload_file takes path as string.

@api_view(['POST'])
def excel_upload(request):
    excel_file = request.FILES['file']
    file_name = request.POST.get('file_name')
    with open('/tmp'+file_name, 'wb+') as destination:
        for chunk in excel_file.chunks():
            destination.write(chunk)
    upload_to_aws('/tmp' + file_name,'X1excelsheets','s3_file_name')

Update upload_to_aws as following.

client_s3.upload_file('/tmp/name.png', bucket, file_name, ExtraArgs={'ContentType': 'image/png'})

answered Nov 14, 2022 at 18:04

Neelabh Singh

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