我如何能看到什么是在S3桶与boto3?(例如,写一个“ls”)?

做以下事情:

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

返回:

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

我如何看到它的内容?


当前回答

如果你想传递ACCESS和SECRET密钥(你不应该这样做,因为这是不安全的):

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)

其他回答

如果你想传递ACCESS和SECRET密钥(你不应该这样做,因为这是不安全的):

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)

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'])

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。

在上面的注释中对@Hephaeastus的代码进行了少许修改,编写了下面的方法来列出给定路径中的文件夹和对象(文件)。类似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

这将列出给定路径下的所有对象/文件夹。Folder_path可以默认为None, method将列出桶根目录的即时内容。

查看内容的一种方法是:

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