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

下面是一个简单的函数,它返回所有文件的文件名或具有特定类型的文件,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

其他回答

#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')

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)

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)

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')

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。