首先，创建一个s3客户端对象:

s3_client = boto3.client('s3')

接下来，创建一个变量来保存bucket名称和文件夹。注意文件夹名后面的斜杠“/”:

bucket_name = 'my-bucket'
folder = 'some-folder/'

接下来，调用s3_client。List_objects_v2获取文件夹内容对象的元数据:

response = s3_client.list_objects_v2(
  Bucket=bucket_name,
  Prefix=folder
)

最后，使用对象的元数据，您可以通过调用s3_client来获取S3对象。get_object功能:

for object_metadata in response['Contents']:
    object_key = object_metadata['Key']
    response = s3_client.get_object(
        Bucket=bucket_name,
        Key=object_key
    )
    object_body = response['Body'].read()
    print(object_body)

如你所见，字符串格式的对象内容可以通过调用response['Body'].read()来获得。

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()

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'])

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

#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')

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'))