我想使用 python 复制 s3 存储桶中的文件。
例如：我有存储桶名称 = 测试。在存储桶中，我有 2 个文件夹，名称为“dump”和“input”。现在我想使用python将文件从本地目录复制到S3“转储”文件夹......有人可以帮助我吗？
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294199 次
15 回答
116
尝试这个...
import boto
import boto.s3
import sys
from boto.s3.key import Key
AWS_ACCESS_KEY_ID = ''
AWS_SECRET_ACCESS_KEY = ''
bucket_name = AWS_ACCESS_KEY_ID.lower() + '-dump'
conn = boto.connect_s3(AWS_ACCESS_KEY_ID,
AWS_SECRET_ACCESS_KEY)
bucket = conn.create_bucket(bucket_name,
location=boto.s3.connection.Location.DEFAULT)
testfile = "replace this with an actual filename"
print 'Uploading %s to Amazon S3 bucket %s' % \
(testfile, bucket_name)
def percent_cb(complete, total):
sys.stdout.write('.')
sys.stdout.flush()
k = Key(bucket)
k.key = 'my test file'
k.set_contents_from_filename(testfile,
cb=percent_cb, num_cb=10)
[更新] 我不是 python 专家，所以感谢您对 import 语句的提醒。另外，我不建议将凭据放在您自己的源代码中。如果您在 AWS 中运行它，请使用带有实例配置文件的 IAM 凭证 ( http://docs.aws.amazon.com/IAM/latest/UserGuide/id_roles_use_switch-role-ec2_instance-profiles.html )，并在您的开发/测试环境，使用 AdRoll 中的全息图（https://github.com/AdRoll/hologram）
于 2013-02-26T11:04:56.557 回答
87
import boto3
s3 = boto3.resource('s3')
BUCKET = "test"
s3.Bucket(BUCKET).upload_file("your/local/file", "dump/file")
于 2017-11-03T15:17:29.067 回答
50
没必要那么复杂：
s3_connection = boto.connect_s3()
bucket = s3_connection.get_bucket('your bucket name')
key = boto.s3.key.Key(bucket, 'some_file.zip')
with open('some_file.zip') as f:
key.send_file(f)
于 2015-06-29T10:00:18.170 回答
36
我用过这个，实现起来非常简单
import tinys3
conn = tinys3.Connection('S3_ACCESS_KEY','S3_SECRET_KEY',tls=True)
f = open('some_file.zip','rb')
conn.upload('some_file.zip',f,'my_bucket')
于 2014-12-24T08:48:46.093 回答
31
在带有凭据的会话中将文件上传到 s3。
import boto3
session = boto3.Session(
aws_access_key_id='AWS_ACCESS_KEY_ID',
aws_secret_access_key='AWS_SECRET_ACCESS_KEY',
)
s3 = session.resource('s3')
# Filename - File to upload
# Bucket - Bucket to upload to (the top level directory under AWS S3)
# Key - S3 object name (can contain subdirectories). If not specified then file_name is used
s3.meta.client.upload_file(Filename='input_file_path', Bucket='bucket_name', Key='s3_output_key')
于 2019-02-08T15:22:27.053 回答
17
from boto3.s3.transfer import S3Transfer
import boto3
#have all the variables populated which are required below
client = boto3.client('s3', aws_access_key_id=access_key,aws_secret_access_key=secret_key)
transfer = S3Transfer(client)
transfer.upload_file(filepath, bucket_name, folder_name+"/"+filename)
于 2017-01-31T12:34:05.487 回答
13
这是一个三班轮。只需按照boto3 文档中的说明进行操作即可。
import boto3
s3 = boto3.resource(service_name = 's3')
s3.meta.client.upload_file(Filename = 'C:/foo/bar/baz.filetype', Bucket = 'yourbucketname', Key = 'baz.filetype')
一些重要的论点是：
参数：
str) -- 要上传的文件的路径。str) -- 要上传到的存储桶的名称。
str) -- 您要分配给 s3 存储桶中文件的名称。这可能与文件名相同或您选择的不同名称，但文件类型应保持不变。
注意：我假设您已按照 boto3 文档~\.aws中的最佳配置实践中的建议将凭据保存在文件夹中。
于 2018-11-29T00:20:09.540 回答
12
这也将起作用：
import os
import boto
import boto.s3.connection
from boto.s3.key import Key
try:
conn = boto.s3.connect_to_region('us-east-1',
aws_access_key_id = 'AWS-Access-Key',
aws_secret_access_key = 'AWS-Secrete-Key',
# host = 's3-website-us-east-1.amazonaws.com',
# is_secure=True, # uncomment if you are not using ssl
calling_format = boto.s3.connection.OrdinaryCallingFormat(),
)
bucket = conn.get_bucket('YourBucketName')
key_name = 'FileToUpload'
path = 'images/holiday' #Directory Under which file should get upload
full_key_name = os.path.join(path, key_name)
k = bucket.new_key(full_key_name)
k.set_contents_from_filename(key_name)
except Exception,e:
print str(e)
print "error"
于 2017-02-01T08:19:25.987 回答
6
使用 boto3
import logging
import boto3
from botocore.exceptions import ClientError
def upload_file(file_name, bucket, object_name=None):
"""Upload a file to an S3 bucket
:param file_name: File to upload
:param bucket: Bucket to upload to
:param object_name: S3 object name. If not specified then file_name is used
:return: True if file was uploaded, else False
"""
# If S3 object_name was not specified, use file_name
if object_name is None:
object_name = file_name
# Upload the file
s3_client = boto3.client('s3')
try:
response = s3_client.upload_file(file_name, bucket, object_name)
except ClientError as e:
logging.error(e)
return False
return True
更多信息：- https://boto3.amazonaws.com/v1/documentation/api/latest/guide/s3-uploading-files.html
于 2019-08-22T11:10:50.707 回答
5
import boto
from boto.s3.key import Key
AWS_ACCESS_KEY_ID = ''
AWS_SECRET_ACCESS_KEY = ''
END_POINT = '' # eg. us-east-1
S3_HOST = '' # eg. s3.us-east-1.amazonaws.com
BUCKET_NAME = 'test'
FILENAME = 'upload.txt'
UPLOADED_FILENAME = 'dumps/upload.txt'
# include folders in file path. If it doesn't exist, it will be created
s3 = boto.s3.connect_to_region(END_POINT,
aws_access_key_id=AWS_ACCESS_KEY_ID,
aws_secret_access_key=AWS_SECRET_ACCESS_KEY,
host=S3_HOST)
bucket = s3.get_bucket(BUCKET_NAME)
k = Key(bucket)
k.key = UPLOADED_FILENAME
k.set_contents_from_filename(FILENAME)
于 2017-03-02T13:23:42.570 回答
2
对于上传文件夹示例，如下代码和 S3 文件夹图片
import boto
import boto.s3
import boto.s3.connection
import os.path
import sys
# Fill in info on data to upload
# destination bucket name
bucket_name = 'willie20181121'
# source directory
sourceDir = '/home/willie/Desktop/x/' #Linux Path
# destination directory name (on s3)
destDir = '/test1/' #S3 Path
#max size in bytes before uploading in parts. between 1 and 5 GB recommended
MAX_SIZE = 20 * 1000 * 1000
#size of parts when uploading in parts
PART_SIZE = 6 * 1000 * 1000
access_key = 'MPBVAQ*******IT****'
secret_key = '11t63yDV***********HgUcgMOSN*****'
conn = boto.connect_s3(
aws_access_key_id = access_key,
aws_secret_access_key = secret_key,
host = '******.org.tw',
is_secure=False, # uncomment if you are not using ssl
calling_format = boto.s3.connection.OrdinaryCallingFormat(),
)
bucket = conn.create_bucket(bucket_name,
location=boto.s3.connection.Location.DEFAULT)
uploadFileNames = []
for (sourceDir, dirname, filename) in os.walk(sourceDir):
uploadFileNames.extend(filename)
break
def percent_cb(complete, total):
sys.stdout.write('.')
sys.stdout.flush()
for filename in uploadFileNames:
sourcepath = os.path.join(sourceDir + filename)
destpath = os.path.join(destDir, filename)
print ('Uploading %s to Amazon S3 bucket %s' % \
(sourcepath, bucket_name))
filesize = os.path.getsize(sourcepath)
if filesize > MAX_SIZE:
print ("multipart upload")
mp = bucket.initiate_multipart_upload(destpath)
fp = open(sourcepath,'rb')
fp_num = 0
while (fp.tell() < filesize):
fp_num += 1
print ("uploading part %i" %fp_num)
mp.upload_part_from_file(fp, fp_num, cb=percent_cb, num_cb=10, size=PART_SIZE)
mp.complete_upload()
else:
print ("singlepart upload")
k = boto.s3.key.Key(bucket)
k.key = destpath
k.set_contents_from_filename(sourcepath,
cb=percent_cb, num_cb=10)
PS：更多参考网址
于 2018-12-03T08:23:35.557 回答
2
如果您的系统上安装了aws 命令行界面，则可以使用 pythons子进程库。例如：
import subprocess
def copy_file_to_s3(source: str, target: str, bucket: str):
subprocess.run(["aws", "s3" , "cp", source, f"s3://{bucket}/{target}"])
同样，您可以将该逻辑用于各种 AWS 客户端操作，例如下载或列出文件等。也可以获取返回值。这样就不需要导入boto3了。我想它的使用不是这样的，但在实践中我觉得这样很方便。这样，您还可以获得控制台中显示的上传状态 - 例如：
Completed 3.5 GiB/3.5 GiB (242.8 MiB/s) with 1 file(s) remaining
要根据您的意愿修改该方法，我建议您查看子流程参考以及AWS Cli 参考。
注意：这是我对类似问题的回答的副本。
于 2020-11-06T11:41:40.570 回答
1
我有一些东西在我看来有更多的顺序：
import boto3
from pprint import pprint
from botocore.exceptions import NoCredentialsError
class S3(object):
BUCKET = "test"
connection = None
def __init__(self):
try:
vars = get_s3_credentials("aws")
self.connection = boto3.resource('s3', 'aws_access_key_id',
'aws_secret_access_key')
except(Exception) as error:
print(error)
self.connection = None
def upload_file(self, file_to_upload_path, file_name):
if file_to_upload is None or file_name is None: return False
try:
pprint(file_to_upload)
file_name = "your-folder-inside-s3/{0}".format(file_name)
self.connection.Bucket(self.BUCKET).upload_file(file_to_upload_path,
file_name)
print("Upload Successful")
return True
except FileNotFoundError:
print("The file was not found")
return False
except NoCredentialsError:
print("Credentials not available")
return False
这里有三个重要的变量，BUCKET const、file_to_upload和file_name
BUCKET: 是您的 S3 存储桶的名称
file_to_upload_path: 必须是您要上传的文件的路径
file_name: 是您的存储桶中的结果文件和路径（这是您添加文件夹或其他任何内容的位置）
有很多方法，但您可以在这样的另一个脚本中重用此代码
import S3
def some_function():
S3.S3().upload_file(path_to_file, final_file_name)
于 2020-06-29T23:31:09.743 回答
0
您还应该提及内容类型以省略文件访问问题。
import os
image='fly.png'
s3_filestore_path = 'images/fly.png'
filename, file_extension = os.path.splitext(image)
content_type_dict={".png":"image/png",".html":"text/html",
".css":"text/css",".js":"application/javascript",
".jpg":"image/png",".gif":"image/gif",
".jpeg":"image/jpeg"}
content_type=content_type_dict[file_extension]
s3 = boto3.client('s3', config=boto3.session.Config(signature_version='s3v4'),
region_name='ap-south-1',
aws_access_key_id=S3_KEY,
aws_secret_access_key=S3_SECRET)
s3.put_object(Body=image, Bucket=S3_BUCKET, Key=s3_filestore_path, ContentType=content_type)
于 2020-10-01T06:56:57.500 回答
0
xmlstr = etree.tostring(listings, encoding='utf8', method='xml')
conn = boto.connect_s3(
aws_access_key_id = access_key,
aws_secret_access_key = secret_key,
# host = '<bucketName>.s3.amazonaws.com',
host = 'bycket.s3.amazonaws.com',
#is_secure=False, # uncomment if you are not using ssl
calling_format = boto.s3.connection.OrdinaryCallingFormat(),
)
conn.auth_region_name = 'us-west-1'
bucket = conn.get_bucket('resources', validate=False)
key= bucket.get_key('filename.txt')
key.set_contents_from_string("SAMPLE TEXT")
key.set_canned_acl('public-read')
于 2018-11-13T19:59:12.173 回答