如何通过boto3
查看S3中存储桶中的内容? (即做一个"ls"
)?
执行以下操作:
import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket('some/path/')
返回:
s3.Bucket(name='some/path/')
如何查看其内容?
答案 0 :(得分:149)
查看内容的一种方法是:
<div>Click and hold me...</div>
<div>Click and hold me...</div>
<div>Click and hold me...</div>
答案 1 :(得分:71)
这类似于'ls',但它没有考虑前缀文件夹约定,并将列出存储桶中的对象。这取决于读者过滤掉作为密钥名称一部分的前缀。
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'])
答案 2 :(得分:21)
我假设您已单独配置身份验证。
import boto3
s3 = boto3.resource('s3')
my_bucket = s3.Bucket('bucket_name')
for file in my_bucket.objects.all():
print file.key
答案 3 :(得分:21)
如果你想传递ACCESS和SECRET键(你不应该这样做,因为它不安全):
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)
答案 4 :(得分:12)
为了处理大型键列表(即,当目录列表大于1000个项目时),我使用以下代码来累加具有多个列表的键值(即文件名)(感谢上面的第一行Amelio)。代码适用于python3:
from boto3 import client
bucket_name = "my_bucket"
prefix = "my_key/sub_key/lots_o_files"
s3_conn = client('s3') # type: BaseClient ## again assumes boto.cfg setup, assume AWS S3
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Prefix=prefix, Delimiter = "/")
if 'Contents' not in s3_result:
#print(s3_result)
return []
file_list = []
for key in s3_result['Contents']:
file_list.append(key['Key'])
print(f"List count = {len(file_list)}")
while s3_result['IsTruncated']:
continuation_key = s3_result['NextContinuationToken']
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Prefix=prefix, Delimiter="/", ContinuationToken=continuation_key)
for key in s3_result['Contents']:
file_list.append(key['Key'])
print(f"List count = {len(file_list)}")
return file_list
答案 5 :(得分:6)
我的s3 keys
utility function本质上是@Hephaestus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[1:] if prefix.startswith(delimiter) else prefix
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']
在我的测试中(boto3 1.9.84),它比等效(但更简单)的代码快得多:
import boto3
def keys(bucket_name, prefix='/', delimiter='/'):
prefix = prefix[1:] if prefix.startswith(delimiter) else prefix
bucket = boto3.resource('s3').Bucket(bucket_name)
return (_.key for _ in bucket.objects.filter(Prefix=prefix))
与S3 guarantees UTF-8 binary sorted results一样,start_after
优化已添加到第一个函数中。
答案 6 :(得分:5)
一种更简约的方式,而不是通过for循环迭代,你也可以打印包含S3存储桶中所有文件的原始对象:
session = Session(aws_access_key_id=aws_access_key_id,aws_secret_access_key=aws_secret_access_key)
s3 = session.resource('s3')
bucket = s3.Bucket('bucket_name')
files_in_s3 = bucket.objects.all()
#you can print this iterable with print(list(files_in_s3))
答案 7 :(得分:2)
对象摘要:
ObjectSummary附带有两个标识符:
AWS S3文档中有关对象键的更多信息:
对象键:
创建对象时,请指定键名,该键名唯一标识存储桶中的对象。例如,在Amazon S3控制台(请参阅AWS管理控制台)中,突出显示存储桶时,将显示存储桶中的对象列表。这些名称是对象键。密钥的名称是一系列Unicode字符,其UTF-8编码最长为1024个字节。
Amazon S3数据模型是一个平面结构:创建一个存储桶,该存储桶存储对象。没有子桶或子文件夹的层次结构;但是,您可以像Amazon S3控制台一样使用键名前缀和定界符来推断逻辑层次结构。 Amazon S3控制台支持文件夹的概念。假设您的存储桶(由管理员创建)具有四个带有以下对象键的对象:
Development / Projects1.xls
财务/声明1.pdf
私人/taxdocument.pdf
s3-dg.pdf
参考:
下面是一些示例代码,演示了如何获取存储桶名称和对象密钥。
示例:
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()
答案 8 :(得分:2)
这是一个简单的函数,可向您返回所有文件或具有某些类型(例如“ json”,“ jpg”)的文件的文件名。
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
答案 9 :(得分:1)
我只是这样做,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
答案 10 :(得分:1)
Content-Type
答案 11 :(得分:1)
在上面的注释之一中对@Hephaeastus的代码进行了很少的修改,编写了以下方法来列出给定路径中的文件夹和对象(文件)。与s3 ls命令类似。
from boto3 import session
def s3_ls(profile=None, bucket_name=None, folder_path=None):
folders=[]
files=[]
result=dict()
bucket_name = bucket_name
prefix= folder_path
session = boto3.Session(profile_name=profile)
s3_conn = session.client('s3')
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Delimiter = "/", Prefix=prefix)
if 'Contents' not in s3_result and 'CommonPrefixes' not in s3_result:
return []
if s3_result.get('CommonPrefixes'):
for folder in s3_result['CommonPrefixes']:
folders.append(folder.get('Prefix'))
if s3_result.get('Contents'):
for key in s3_result['Contents']:
files.append(key['Key'])
while s3_result['IsTruncated']:
continuation_key = s3_result['NextContinuationToken']
s3_result = s3_conn.list_objects_v2(Bucket=bucket_name, Delimiter="/", ContinuationToken=continuation_key, Prefix=prefix)
if s3_result.get('CommonPrefixes'):
for folder in s3_result['CommonPrefixes']:
folders.append(folder.get('Prefix'))
if s3_result.get('Contents'):
for key in s3_result['Contents']:
files.append(key['Key'])
if folders:
result['folders']=sorted(folders)
if files:
result['files']=sorted(files)
return result
这将列出给定路径中的所有对象/文件夹。默认情况下,Folder_path可以保留为None,方法将列出存储桶根的立即内容。
答案 12 :(得分:1)
因此,您要在boto3中要求aws s3 ls
的等价物。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。它仅列出所有顶级文件夹。令人惊讶的是,如此简单的操作有多么困难。
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'))
答案 13 :(得分:1)
我曾经这样做的一种方式:
import boto3
s3 = boto3.resource('s3')
bucket=s3.Bucket("bucket_name")
contents = [_.key for _ in bucket.objects.all() if "subfolders/ifany/" in _.key]
答案 14 :(得分:0)
这是解决方案
import boto3
s3=boto3.resource('s3')
BUCKET_NAME = 'Your S3 Bucket Name'
allFiles = s3.Bucket(BUCKET_NAME).objects.all()
for file in allFiles:
print(file.key)
答案 15 :(得分:0)
也可以按照以下步骤操作:
import React from 'react';
import {Header} from './header.js';
import {Searchbar} from './searchbar.js';
import {Thumbnails}from './Thumbnails.js';
import {connect} from 'react-redux';
import {Select} from './select.js';
import { Pagination } from './pagination.js';
import {Filterbar} from './filterbar.js';
import { popular } from '../redux/action.js';
class App extends React.Component {
componentDidMount(){
console.log("heloo");
popular();
}
render(){
return (
<>
<Header />
<Select {...this.props}/>
<Filterbar/>
<Searchbar {...this.props}/>
<Thumbnails {...this.props}/>
<Pagination {...this.props}/>
</>
);
}
}
function mapStateToProps(state){
return {
state:state
}
}
export default connect(mapStateToProps)(App)
答案 16 :(得分:0)
import boto3
s3 = boto3.resource('s3')
## Bucket to use
my_bucket = s3.Bucket('city-bucket')
## List objects within a given prefix
for obj in my_bucket.objects.filter(Delimiter='/', Prefix='city/'):
print obj.key
输出:
city/pune.csv
city/goa.csv