从lambda函数运行aws cli命令也是一个不错的选择

import subprocess
import logging

logger = logging.getLogger()
logger.setLevel(logging.INFO)

def run_command(command):
    command_list = command.split(' ')

    try:
        logger.info("Running shell command: \"{}\"".format(command))
        result = subprocess.run(command_list, stdout=subprocess.PIPE);
        logger.info("Command output:\n---\n{}\n---".format(result.stdout.decode('UTF-8')))
    except Exception as e:
        logger.error("Exception: {}".format(e))
        return False

    return True

def lambda_handler(event, context):
    run_command('/opt/aws s3 ls s3://bucket-name')

这是解决方案

import boto3

s3=boto3.resource('s3')
BUCKET_NAME = 'Your S3 Bucket Name'
allFiles = s3.Bucket(BUCKET_NAME).objects.all()
for file in allFiles:
    print(file.key)

ObjectSummary:

有两个标识符附加到ObjectSummary:

bucket_name 关键

boto3 S3: ObjectSummary

有关AWS S3文档中的对象键的更多信息:

Object Keys: When you create an object, you specify the key name, which uniquely identifies the object in the bucket. For example, in the Amazon S3 console (see AWS Management Console), when you highlight a bucket, a list of objects in your bucket appears. These names are the object keys. The name for a key is a sequence of Unicode characters whose UTF-8 encoding is at most 1024 bytes long. The Amazon S3 data model is a flat structure: you create a bucket, and the bucket stores objects. There is no hierarchy of subbuckets or subfolders; however, you can infer logical hierarchy using key name prefixes and delimiters as the Amazon S3 console does. The Amazon S3 console supports a concept of folders. Suppose that your bucket (admin-created) has four objects with the following object keys: Development/Projects1.xls Finance/statement1.pdf Private/taxdocument.pdf s3-dg.pdf Reference: AWS S3: Object Keys

下面是一些示例代码，演示如何获取桶名和对象键。

例子:

import boto3
from pprint import pprint

def main():

    def enumerate_s3():
        s3 = boto3.resource('s3')
        for bucket in s3.buckets.all():
             print("Name: {}".format(bucket.name))
             print("Creation Date: {}".format(bucket.creation_date))
             for object in bucket.objects.all():
                 print("Object: {}".format(object))
                 print("Object bucket_name: {}".format(object.bucket_name))
                 print("Object key: {}".format(object.key))

    enumerate_s3()


if __name__ == '__main__':
    main()

我只是这样做的，包括身份验证方法:

s3_client = boto3.client(
                's3',
                aws_access_key_id='access_key',
                aws_secret_access_key='access_key_secret',
                config=boto3.session.Config(signature_version='s3v4'),
                region_name='region'
            )

response = s3_client.list_objects(Bucket='bucket_name', Prefix=key)
if ('Contents' in response):
    # Object / key exists!
    return True
else:
    # Object / key DOES NOT exist!
    return False

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))

我的s3键实用函数本质上是@Hephaestus的答案的优化版本:

import boto3


s3_paginator = boto3.client('s3').get_paginator('list_objects_v2')


def keys(bucket_name, prefix='/', delimiter='/', start_after=''):
    prefix = prefix.lstrip(delimiter)
    start_after = (start_after or prefix) if prefix.endswith(delimiter) else start_after
    for page in s3_paginator.paginate(Bucket=bucket_name, Prefix=prefix, StartAfter=start_after):
        for content in page.get('Contents', ()):
            yield content['Key']

在我的测试(boto3 1.9.84)中，它比等效的(但更简单)代码要快得多:

import boto3


def keys(bucket_name, prefix='/', delimiter='/'):
    prefix = prefix.lstrip(delimiter)
    bucket = boto3.resource('s3').Bucket(bucket_name)
    return (_.key for _ in bucket.objects.filter(Prefix=prefix))

由于S3保证UTF-8二进制排序结果，因此在第一个函数中添加了start_after优化。