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?
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
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。
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)
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'])
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