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

#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')

其他回答

为了处理大型键列表(即当目录列表大于1000项时),我使用以下代码将多个列表中的键值(即文件名)累积起来(感谢上面的阿梅里奥的第一行)。代码是针对python3的:

    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

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'))

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'])

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)