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
我只是这样做的,包括身份验证方法:
s3_client = boto3.client(
's3',
aws_access_key_id='access_key',
aws_secret_access_key='access_key_secret',
config=boto3.session.Config(signature_version='s3v4'),
region_name='region'
)
response = s3_client.list_objects(Bucket='bucket_name', Prefix=key)
if ('Contents' in response):
# Object / key exists!
return True
else:
# Object / key DOES NOT exist!
return False
其他回答
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'))
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)
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。
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')