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

使用cloudpathlib

cloudpathlib提供了一个方便的包装器,这样您就可以使用简单的pathlib API与AWS S3(以及Azure blob存储、GCS等)进行交互。你可以用pip install "cloudpathlib[s3]"来安装。

像pathlib一样,你可以使用glob或iterdir来列出目录的内容。

下面是一个带有公共AWS S3桶的示例,您可以复制并过去运行该桶。

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')]

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

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'))

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

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()

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)