python - 类型错误 : expected string or bytes-like object - how to copy files to an S3 bucket using the code provided in AWS's documentation

Question

https://boto3.amazonaws.com/v1/documentation/api/latest/guide/s3-uploading-files.html

使用文档中提供的代码，我尝试迭代parent_dir，如果有zip 文件，那么我想将其复制到我的S3 存储桶。

我都尝试过

    try:
        response = s3_client.upload_file(file_name, bucket, object_name)
    except ClientError as e:
        logging.error(e)
        return False
    return True

和

s3 = boto3.client('s3')
with open("FILE_NAME", "rb") as f:
    s3.upload_fileobj(f, "BUCKET_NAME", "OBJECT_NAME")

但是他们都给出了同样的错误。

    s3_client = boto3.client(
        's3',
        aws_access_key_id='MY_KEY_ID',
        aws_secret_access_key='MY_ACCESS_KEY'
    )

    session = boto3.Session(
        aws_access_key_id='MY_KEY_ID',
        aws_secret_access_key='MY_ACCESS_KEY',
    )

    s3 = session.resource('s3')
    bucket = s3.Bucket('MY_URL')

    for file in os.listdir(parent_dir):
        if object_name is None:
            object_name = file

        if file.endswith('.zip'):
            with open(file, "rb") as f:
                s3_client.upload_fileobj(f, bucket, object_name)

类型错误:预期的字符串或类似字节的对象

Answer 1

根据[AmazonAWS.Boto3]: S3.Client - upload_fileobj(Fileobj, Bucket, Key, ExtraArgs=None, Callback=None, Config=None) ，第二个^nd和第三个^rd参数(Bucket和Key)必须是字符串。

但是您将作为第二个^nd参数传递:

bucket = s3.Bucket('MY_URL')

这不行。使其成为纯字符串(更好的是，重命名它):

bucket_name = "MY_URL"

并将其传递给upload_fileobj，您应该可以解决这个问题。