使用Boto3将文件上载到带有前缀的S3存储桶

时间:2016-05-04 02:06:13

标签: python amazon-s3 boto3

我正在尝试将文件上传到S3存储桶,但我无法访问存储桶的根级别,而是需要将其上传到某个前缀。以下代码:

import boto3
s3 = boto3.resource('s3')
open('/tmp/hello.txt', 'w+').write('Hello, world!')
s3_client.upload_file('/tmp/hello.txt', bucket_name, prefix+'hello-remote.txt')

给我一​​个错误:

An error occurred (AccessDenied) when calling the PutObject operation: Access Denied: ClientError Traceback (most recent call last): File "/var/task/tracker.py", line 1009, in testHandler s3_client.upload_file('/tmp/hello.txt', bucket_name, prefix+'hello-remote.txt') File "/var/runtime/boto3/s3/inject.py", line 71, in upload_file extra_args=ExtraArgs, callback=Callback) File "/var/runtime/boto3/s3/transfer.py", line 641, in upload_file self._put_object(filename, bucket, key, callback, extra_args) File "/var/runtime/boto3/s3/transfer.py", line 651, in _put_object **extra_args) File "/var/runtime/botocore/client.py", line 228, in _api_call return self._make_api_call(operation_name, kwargs) File "/var/runtime/botocore/client.py", line 492, in _make_api_call raise ClientError(parsed_response, operation_name) ClientError: An error occurred (AccessDenied) when calling the PutObject operation: Access Denied

bucket_name的格式为abcd,而prefix的格式为a/b/c/d/。我不确定错误是由于斜线是错误还是在某种程度上你可以在其他地方指定前缀,或者我是否没有写入权限(尽管我认为这样做)

此代码执行时没有任何错误:

for object in output_bucket.objects.filter(Prefix=prefix):
    print(object.key)

虽然没有输出,因为存储桶为空。

5 个答案:

答案 0 :(得分:6)

原来我需要SSE:

transfer = S3Transfer(s3_client)
transfer.upload_file('/tmp/hello.txt', bucket_name, prefix+'hello-remote.txt', extra_args={'ServerSideEncryption': "AES256"})

答案 1 :(得分:2)

我假设您完成了所有这些设置:

  1. AWS Access Key ID和Secret Key设置(通常存储在~/.aws/credentials
  2. 您可以访问S3,并且您知道您的存储桶名称&前缀(子目录)
  3. 根据Boto3 S3 upload_file documentation,您应该按照以下方式上传您的上传:

    upload_file(Filename, Bucket, Key, ExtraArgs=None, Callback=None, Config=None)

    import boto3
    s3 = boto3.resource('s3')
    s3.meta.client.upload_file('/tmp/hello.txt', 'mybucket', 'hello.txt')
    

    这里要注意的关键是s3.meta.client。别忘了 - 它对我有用!

    我希望有所帮助。

答案 2 :(得分:2)

以下是John Adjei回答的替代方法。这也取自Boto3 S3 upload_file documentation。因为客户端是低级别的(低抽象度/更接近机器代码),所以它可以提高性能-尤其是在处理大数据时。

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

答案 3 :(得分:0)

import boto3

s3 = boto3.resource('s3')
s3.meta.client.upload_file( 'csv1.csv', "bucketname", "prefixna/csv1.csv")

答案 4 :(得分:0)

这是我的答案:

import boto3

s3_client = boto3.client(service_name='s3', region_name='ap-southeast-1',
                         aws_access_key_id='AWS_ACCESS_KEY_ID',
                         aws_secret_access_key='AWS_SECRET_ACCESS_KEY')

dest_bucket = 'data-lake'
dest_prefix = 'datamart/my_file_name/'

file_name = 'my_file_name'+ '.parquet'

s3.meta.client.delete_object(Bucket=dest_bucket,Key=dest_prefix + file_name)