如何使用博托文件上传到目录中S3桶

Question

我想用Python中的文件复制在S3存储桶。

例如：我有水桶名=测试。而在斗，我有2个文件夹命名为“垃圾”和“输入”。现在我想使用Python从本地目录中的文件复制到S3“垃圾”文件夹......谁能帮助我？

Answer 1

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pythonist，所以感谢抬起头对import语句。另外，我建议不要将凭据自己的源代码内。如果您正在运行这里面AWS使用与实例配置文件（http://docs.aws.amazon.com/IAM/latest/UserGuide/id_roles_use_switch-role-ec2_instance-profiles.html）IAM凭证，并保持相同的行为在你的开发/测试环境中，使用这样的全息图将AdRoll（https://github.com/AdRoll/hologram）

Answer 2

用于上载的文件夹例如如下面的代码和S3文件夹图像

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URL

Answer 3

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 4

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

欲了解更多： - https://boto3.amazonaws.com/v1/documentation/api/latest/guide/s3-uploading-files.html

Answer 5

没有必要弄得这么复杂：

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 6

我用这个，这是很容易实现

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 7

import boto3

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

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

Answer 8

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 9

这也将工作：

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 10

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 11

上传文件与证书的会话中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')

Answer 12

这是一个三个衬垫。只需按照boto3 documentation的说明。

import boto3
s3 = boto3.resource(service_name = 's3')
s3.meta.client.upload_file(Filename = 'C:/foo/bar/baz.filetype', Bucket = 'yourbucketname', Key = 'baz.filetype')

一些重要的参数是：

参数：

文件名（str） - 该文件的路径进行上传。

桶（str） - 斗上传到的名称。

密钥（str） - 的，你要分配给您的文件在您的S3存储桶的名称。这可能是一样的文件的名称或您选择的不同的名字，但文件类型应保持不变。 注：我假设你已经保存在一个文件夹~\.aws您的凭据作为best configuration practices in the boto3 documentation建议。