Python 使用 Boto3 将文件上传到带有前缀的 S3 存储桶
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Uploading a file to a S3 bucket with a prefix using Boto3
提问by foxes
I am attempting to upload a file into a S3 bucket, but I don't have access to the root level of the bucket and I need to upload it to a certain prefix instead. The following code:
我正在尝试将文件上传到 S3 存储桶，但我无权访问存储桶的根级别，我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')
Gives me an error:
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
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_nameis in the format abcdwhile prefixis in the format a/b/c/d/. I'm not sure if the error is due to the slashes being wrong or if there's some way you can specify the prefix elsewhere, or if I don't have write permissions (although I supposedly do).
bucket_name在格式中abcd而prefix在格式中a/b/c/d/。我不确定该错误是否是由于斜杠错误造成的，或者是否有某种方法可以在其他地方指定前缀，或者我没有写权限（尽管我应该这样做）。
This code executes without any errors:
此代码执行时没有任何错误：
for object in output_bucket.objects.filter(Prefix=prefix):
print(object.key)
Although there is no output as the bucket is empty.
虽然没有输出，因为桶是空的。
回答by John Adjei
I'm assuming you have all this set up:
我假设你已经设置了所有这些：
- AWS Access Key ID and Secret Key set up (typically stored at
~/.aws/credentials - You have access to S3 and you know your bucket names & prefixes (subdirectories)
- AWS 访问密钥 ID 和秘密密钥设置（通常存储在
~/.aws/credentials - 您有权访问 S3 并且知道您的存储桶名称和前缀（子目录）
According to the Boto3 S3 upload_filedocumentation, you should upload your upload like this:
根据Boto3 S3upload_file文档，您应该像这样上传您的上传：
upload_file(Filename, Bucket, Key, ExtraArgs=None, Callback=None, Config=None)
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')
The key to note here is s3.meta.client. Don't forget that--it worked for me!
这里要注意的关键是s3.meta.client。不要忘记——它对我有用！
I hope that helped.
我希望这有帮助。
回答by foxes
Turns out I needed SSE:
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"})
回答by Ranajit kumar
import boto3
s3 = boto3.resource('s3')
s3.meta.client.upload_file( 'csv1.csv', "bucketname", "prefixna/csv1.csv")
