首先，创建一个s3客户端对象:

s3_client = boto3.client('s3')

接下来，创建一个变量来保存bucket名称和文件夹。注意文件夹名后面的斜杠“/”:

bucket_name = 'my-bucket'
folder = 'some-folder/'

接下来，调用s3_client。List_objects_v2获取文件夹内容对象的元数据:

response = s3_client.list_objects_v2(
  Bucket=bucket_name,
  Prefix=folder
)

最后，使用对象的元数据，您可以通过调用s3_client来获取S3对象。get_object功能:

for object_metadata in response['Contents']:
    object_key = object_metadata['Key']
    response = s3_client.get_object(
        Bucket=bucket_name,
        Key=object_key
    )
    object_body = response['Body'].read()
    print(object_body)

如你所见，字符串格式的对象内容可以通过调用response['Body'].read()来获得。

查看内容的一种方法是:

for my_bucket_object in my_bucket.objects.all():
    print(my_bucket_object)

我假设您已经单独配置了身份验证。

import boto3
s3 = boto3.resource('s3')

my_bucket = s3.Bucket('bucket_name')

for file in my_bucket.objects.all():
    print(file.key)

所以你在boto3中要求等同于aws s3 ls。这将列出所有顶级文件夹和文件。这是我能得到的最接近的结果;它只列出所有顶级文件夹。这么简单的操作居然这么难。

import boto3

def s3_ls():
  s3 = boto3.resource('s3')
  bucket = s3.Bucket('example-bucket')
  result = bucket.meta.client.list_objects(Bucket=bucket.name,
                                           Delimiter='/')
  for o in result.get('CommonPrefixes'):
    print(o.get('Prefix'))

一种更节俭的方法，而不是通过一个for循环来迭代，你也可以只打印原始对象，其中包含S3桶中的所有文件:

session = Session(aws_access_key_id=aws_access_key_id,aws_secret_access_key=aws_secret_access_key)
s3 = session.resource('s3')
bucket = s3.Bucket('bucket_name')

files_in_s3 = bucket.objects.all() 
#you can print this iterable with print(list(files_in_s3))

#To print all filenames in a bucket
import boto3

s3 = boto3.client('s3')

def get_s3_keys(bucket):

    """Get a list of keys in an S3 bucket."""
    resp = s3.list_objects_v2(Bucket=bucket)
    for obj in resp['Contents']:
      files = obj['Key']
    return files

  
filename = get_s3_keys('your_bucket_name')

print(filename)

#To print all filenames in a certain directory in a bucket
import boto3

s3 = boto3.client('s3')

def get_s3_keys(bucket, prefix):

    """Get a list of keys in an S3 bucket."""
    resp = s3.list_objects_v2(Bucket=bucket, Prefix=prefix)
    for obj in resp['Contents']:
      files = obj['Key']
      print(files)
    return files

  
filename = get_s3_keys('your_bucket_name', 'folder_name/sub_folder_name/')

print(filename)

更新: 最简单的方法是使用awswrangler

import awswrangler as wr
wr.s3.list_objects('s3://bucket_name')