Flask Upload Image to S3 without saving it to local file system

iTo achieve that with a FileStorage, I use the method put_object():

from werkzeug import secure_filename@user_api.route('upload-profile-photo', methods=['PUT'])@Auth.auth_requireddef upload_profile_photo():    """    Upload User Profile Photo    """    key = Auth.auth_user()    bucket = 'profile-photos'    content_type = request.mimetype    image_file = request.files['file']    client = boto3.client('s3',                          region_name='sfo2',                          endpoint_url='https://example.xxx.amazonaws.com',                          aws_access_key_id=os.environ['ACCESS_KEY'],                          aws_secret_access_key=os.environ['SECRET_KEY'])    filename = secure_filename(image_file.filename)  # This is convenient to validate your filename, otherwise just use file.filename    client.put_object(Body=image_file,                      Bucket=bucket,                      Key=filename,                      ContentType=content_type)    return custom_response({'message': 'image uploaded'}, 200)

Note the call to secure_filename() is optional (you can simply pass image_file.filename), but can be very handy to validate the filename. Otherwise it would be nice to add some exception handlings, but the rough idea is here: no need to open() the file (that would need to be stored locally).

I encourage to have a look at the documentation here, to understand the difference with upload_fileobj()

  import boto3     s3 = boto3.client(            "s3",            aws_access_key_id= "*************",            aws_secret_access_key="**********"        )   @login_blueprint.route('/uploadfile', methods=['POST'])         file= request.files['file']                 try:                    filename = secure_filename(file.filename)                    acl="public-read"                    s3.upload_fileobj(                        file,                        'bucket-name',                        file.filename,                        ExtraArgs={                            "ACL": acl,                            "ContentType":  file.content_type                        }                    )                 except Exception as e:                    resp = jsonify(e)                    resp.status_code =200                    return resp