305

如何查看 S3 中存储桶内的内容boto3?(即做一个"ls")?

执行以下操作:

import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket('some/path/')

返回:

s3.Bucket(name='some/path/')

如何查看其内容?

4

19 回答 19

345

查看内容的一种方法是:

for my_bucket_object in my_bucket.objects.all():
    print(my_bucket_object)
于 2015-05-15T00:17:48.053 回答
140

这类似于“ls”,但它不考虑前缀文件夹约定,并将列出存储桶中的对象。由读者自行过滤掉作为键名一部分的前缀。

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'])
于 2015-05-15T14:45:35.440 回答
85

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)
于 2017-04-05T04:04:03.720 回答
58

我的s3keys实用程序函数本质上是@Hephaestus 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[1:] if prefix.startswith(delimiter) else prefix
    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(但更简单)的代码要快得多:

import boto3


def keys(bucket_name, prefix='/', delimiter='/'):
    prefix = prefix[1:] if prefix.startswith(delimiter) else prefix
    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start_after因此在第一个函数中添加了优化。

于 2019-01-03T00:19:33.183 回答
49

为了处理大型键列表(即当目录列表大于 1000 项时),我使用以下代码来累积具有多个列表的键值(即文件名)(感谢上面的 Amelio 第一行)。代码适用于 python3:

    from boto3  import client
    bucket_name = "my_bucket"
    prefix      = "my_key/sub_key/lots_o_files"

    s3_conn   = client('s3')  # type: BaseClient  ## again assumes boto.cfg setup, assume AWS S3
    s3_result =  s3_conn.list_objects_v2(Bucket=bucket_name, Prefix=prefix, Delimiter = "/")

    if 'Contents' not in s3_result:
        #print(s3_result)
        return []

    file_list = []
    for key in s3_result['Contents']:
        file_list.append(key['Key'])
    print(f"List count = {len(file_list)}")

    while s3_result['IsTruncated']:
        continuation_key = s3_result['NextContinuationToken']
        s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Prefix=prefix, Delimiter="/", ContinuationToken=continuation_key)
        for key in s3_result['Contents']:
            file_list.append(key['Key'])
        print(f"List count = {len(file_list)}")
    return file_list
于 2018-11-22T01:17:47.967 回答
37

如果你想传递 ACCESS 和 SECRET 密钥(你不应该这样做,因为它不安全):

from boto3.session import Session

ACCESS_KEY='your_access_key'
SECRET_KEY='your_secret_key'

session = Session(aws_access_key_id=ACCESS_KEY,
                  aws_secret_access_key=SECRET_KEY)
s3 = session.resource('s3')
your_bucket = s3.Bucket('your_bucket')

for s3_file in your_bucket.objects.all():
    print(s3_file.key)
于 2017-04-07T13:16:47.883 回答
9

一种更简洁的方式,而不是通过 for 循环进行迭代,您还可以只打印包含 S3 存储桶内所有文件的原始对象:

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))
于 2017-06-24T07:14:41.683 回答
7
#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')
于 2019-11-18T22:26:51.867 回答
6

对象总结:

有两个标识符附加到 ObjectSummary:

  • 桶名
  • 钥匙

boto3 S3:对象摘要

AWS S3 文档中有关对象键的更多信息:

对象键:

创建对象时,指定键名,唯一标识存储桶中的对象。例如,在 Amazon S3 控制台(请参阅 AWS 管理控制台)中,当您突出显示存储桶时,会显示存储桶中的对象列表。这些名称是对象键。键的名称是 Unicode 字符序列,其 UTF-8 编码长度最多为 1024 个字节。

Amazon S3 数据模型是一个平面结构:您创建一个存储桶,存储桶存储对象。没有子桶或子文件夹的层次结构;但是,您可以像 Amazon S3 控制台那样使用键名前缀和分隔符来推断逻辑层次结构。Amazon S3 控制台支持文件夹的概念。假设您的存储桶(管理员创建)有四个对象,对象键如下:

开发/Projects1.xls

财务/statement1.pdf

私人/taxdocument.pdf

s3-dg.pdf

参考:

AWS S3:对象键

下面是一些示例代码,演示了如何获取存储桶名称和对象键。

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()
于 2018-10-16T19:26:13.370 回答
6

所以你要求相当于aws s3 lsin boto3. 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;它只列出所有顶级文件夹。令人惊讶的是,如此简单的操作是多么困难。

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'))
于 2020-07-03T13:04:35.630 回答
6
import boto3
s3 = boto3.resource('s3')

## Bucket to use
my_bucket = s3.Bucket('city-bucket')

## List objects within a given prefix
for obj in my_bucket.objects.filter(Delimiter='/', Prefix='city/'):
  print obj.key

输出:

city/pune.csv
city/goa.csv
于 2021-07-27T07:04:04.883 回答
5

这是一个简单的函数,它返回所有文件的文件名或具有某些类型的文件,例如“json”、“jpg”。

def get_file_list_s3(bucket, prefix="", file_extension=None):
            """Return the list of all file paths (prefix + file name) with certain type or all
            Parameters
            ----------
            bucket: str
                The name of the bucket. For example, if your bucket is "s3://my_bucket" then it should be "my_bucket"
            prefix: str
                The full path to the the 'folder' of the files (objects). For example, if your files are in 
                s3://my_bucket/recipes/deserts then it should be "recipes/deserts". Default : ""
            file_extension: str
                The type of the files. If you want all, just leave it None. If you only want "json" files then it
                should be "json". Default: None       
            Return
            ------
            file_names: list
                The list of file names including the prefix
            """
            import boto3
            s3 = boto3.resource('s3')
            my_bucket = s3.Bucket(bucket)
            file_objs =  my_bucket.objects.filter(Prefix=prefix).all()
            file_names = [file_obj.key for file_obj in file_objs if file_extension is not None and file_obj.key.split(".")[-1] == file_extension]
            return file_names
于 2020-07-17T09:46:28.390 回答
5

我曾经这样做的一种方法:

import boto3
s3 = boto3.resource('s3')
bucket=s3.Bucket("bucket_name")
contents = [_.key for _ in bucket.objects.all() if "subfolders/ifany/" in _.key]
于 2020-11-20T07:57:47.797 回答
3

我就是这样做的,包括认证方法:

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
于 2018-09-30T19:21:10.017 回答
2

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)
于 2020-06-07T07:05:09.423 回答
1

在上述评论之一中对@Hephaeastus 的代码几乎没有修改,编写了以下方法来列出给定路径中的文件夹和对象(文件)。工作原理类似于 s3 ls 命令。

from boto3 import session

def s3_ls(profile=None, bucket_name=None, folder_path=None):
    folders=[]
    files=[]
    result=dict()
    bucket_name = bucket_name
    prefix= folder_path
    session = boto3.Session(profile_name=profile)
    s3_conn   = session.client('s3')
    s3_result =  s3_conn.list_objects_v2(Bucket=bucket_name, Delimiter = "/", Prefix=prefix)
    if 'Contents' not in s3_result and 'CommonPrefixes' not in s3_result:
        return []

    if s3_result.get('CommonPrefixes'):
        for folder in s3_result['CommonPrefixes']:
            folders.append(folder.get('Prefix'))

    if s3_result.get('Contents'):
        for key in s3_result['Contents']:
            files.append(key['Key'])

    while s3_result['IsTruncated']:
        continuation_key = s3_result['NextContinuationToken']
        s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Delimiter="/", ContinuationToken=continuation_key, Prefix=prefix)
        if s3_result.get('CommonPrefixes'):
            for folder in s3_result['CommonPrefixes']:
                folders.append(folder.get('Prefix'))
        if s3_result.get('Contents'):
            for key in s3_result['Contents']:
                files.append(key['Key'])

    if folders:
        result['folders']=sorted(folders)
    if files:
        result['files']=sorted(files)
    return result

这列出了给定路径中的所有对象/文件夹。默认情况下,Folder_path 可以保留为 None 并且方法将列出存储桶根目录的直接内容。

于 2020-02-24T03:36:05.107 回答
1

也可以按如下方式进行:

csv_files = s3.list_objects_v2(s3_bucket_path)
    for obj in csv_files['Contents']:
        key = obj['Key']
于 2020-07-02T12:17:29.590 回答
1

一个不错的选择也可能是从 lambda 函数运行 aws cli 命令

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')
于 2021-08-24T20:31:29.973 回答
0

使用cloudpathlib

cloudpathlib提供了一个方便的包装器,以便您可以使用简单的pathlibAPI 与 AWS S3(以及 Azure blob 存储、GCS 等)进行交互。您可以使用pip install "cloudpathlib[s3]".

pathlib您一样可以使用globiterdir列出目录的内容。

这是一个包含公共 AWS S3 存储桶的示例,您可以复制并粘贴该存储桶以运行。

from cloudpathlib import CloudPath

s3_path = CloudPath("s3://ladi/Images/FEMA_CAP/2020/70349")

# list items with glob
list(
    s3_path.glob("*")
)[:3]
#> [ S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0001_5a63d42e-27c6-448a-84f1-bfc632125b8e.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0002_a89f1b79-786f-4dac-9dcc-609fb1a977b1.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0003_02c30af6-911e-4e01-8c24-7644da2b8672.jpg')]

# list items with iterdir
list(
    s3_path.iterdir()
)[:3]
#> [ S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0001_5a63d42e-27c6-448a-84f1-bfc632125b8e.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0002_a89f1b79-786f-4dac-9dcc-609fb1a977b1.jpg'),
#>   S3Path('s3://ladi/Images/FEMA_CAP/2020/70349/DSC_0003_02c30af6-911e-4e01-8c24-7644da2b8672.jpg')]

由reprexlite v0.4.2创建于 2021-05-21 20:38:47 PDT

于 2021-05-22T03:42:08.400 回答