使用Boto3将文件上载到带有前缀的S3存储桶

Question

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是格式abcd时prefix的格式a/b/c/d/.我不确定错误是否是由于斜线错误或者是否有某种方式可以在其他地方指定前缀,或者如果我没有写入权限(尽管我应该这样做).

此代码执行时没有任何错误:

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

虽然铲斗是空的,但没有输出.

Answer 1

import boto3

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

Answer 2

结果我需要SSE:

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

Answer 3

和resource

s3 = boto3.resource('s3')
s3.Bucket('mybucket').upload_file('/tmp/hello.txt', '/detination/s3/path/hello.txt')

和client

s3_client = boto3.client('s3')
s3_client.upload_file('/tmp/hello.txt', 'BUCKET_NAME', '/detination/s3/path/hello.txt',)

Answer 4

我假设您已完成所有这些设置：

1. 设置了AWS Access Key ID和Secret Key（通常存储在 ~/.aws/credentials
2. 您可以访问S3，并且知道您的存储桶名称和前缀（子目录）

根据Boto3 S3 upload_file文档，您应该像这样上载您的上载：

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！

希望对您有所帮助。

Answer 5

这是我的回答：

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)

Answer 6

以下是 John Adjei 答案的替代方案。这也取自Boto3 S3 upload_file 文档。因为客户端是低级的（低抽象/更接近机器代码），它可以提高性能 - 特别是如果您处理大数据。

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