#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')

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()

也可以这样做:

csv_files = s3.list_objects_v2(s3_bucket_path)
    for obj in csv_files['Contents']:
        key = obj['Key']

这类似于'ls'，但它没有考虑到前缀文件夹约定，并将列出bucket中的对象。由读取器来过滤掉作为Key名称一部分的前缀。

在Python 2中:

from boto.s3.connection import S3Connection

conn = S3Connection() # assumes boto.cfg setup
bucket = conn.get_bucket('bucket_name')
for obj in bucket.get_all_keys():
    print(obj.key)

在Python 3中:

from boto3 import client

conn = client('s3')  # again assumes boto.cfg setup, assume AWS S3
for key in conn.list_objects(Bucket='bucket_name')['Contents']:
    print(key['Key'])

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))

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')