Question

我想使用python在s3存储桶中复制一个文件。

Ex：我有桶名=测试。在桶中，我有2个文件夹名称“dump”＆amp; “输入”。现在我想使用python将文件从本地目录复制到S3“dump”文件夹......任何人都可以帮助我吗？

Answer 1

试试这个......

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)

[UPDATE] 我不是pythonist，所以感谢关于import语句的提醒。另外，我不建议在自己的源代码中放置凭据。如果您在AWS内部运行此项，请使用带有实例配置文件（http://docs.aws.amazon.com/IAM/latest/UserGuide/id_roles_use_switch-role-ec2_instance-profiles.html）的IAM凭据，并在开发/测试环境中保持相同的行为，请使用AdRoll中的全息图（https://github.com/AdRoll/hologram）

Answer 2

不需要那么复杂：

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)

Answer 3

我使用了它，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')

https://www.smore.com/labs/tinys3/

Answer 4

import boto3

s3 = boto3.resource('s3')
BUCKET = "test"

s3.Bucket(BUCKET).upload_file("your/local/file", "dump/file")

Answer 5

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)

Answer 6

这也有效：

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"

Answer 7

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)

Answer 8

这是三个衬里。只需按照boto3 documentation上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：

文件名（str）-要上传的文件的路径。

存储桶（str）-要上传到的存储桶的名称。

键（str）-您要分配给s3存储桶中文件的的名称。该名称可以与文件名相同，也可以与您选择的名称不同，但是文件类型应保持不变。

注意：我假设您已按照best configuration practices in the boto3 documentation中的建议将凭据保存在~\.aws文件夹中。

Answer 9

在具有凭据的会话中将文件上传到s3。

_schemas

Answer 10

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：有关更多参考，URL

Answer 11

我觉得有些东西还需要点命令：

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

这里有三个重要变量，即 BUCKET 常量， 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：是存储桶中的结果文件和路径（这是您添加文件夹或其他文件的位置）

有很多方法，但是您可以在这样的另一个脚本中重复使用此代码

import S3

def some_function():
    S3.S3().upload_file(path_to_file, final_file_name)

Answer 12

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')

Answer 13

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

有关更多：- https://boto3.amazonaws.com/v1/documentation/api/latest/guide/s3-uploading-files.html

Answer 14

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)

Answer 15

如果您的系统上安装了aws command line interface，则可以使用pythons subprocess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}"])

类似地，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。还可以获取返回值。这样就无需导入boto3。我想它的用途不是那样的，但实际上，我觉得那样非常方便。这样，您还可以在控制台中显示上载的状态-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方法，建议您参考subprocess参考和AWS Cli reference。

注意：这是我对similar question的回答的副本。

如何使用boto将文件上传到S3存储桶中的目录

15 个答案: