Python 如何使用boto将文件上传到S3存储桶中的目录
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How to upload a file to directory in S3 bucket using boto
提问by Dheeraj Gundra
I want to copy a file in s3 bucket using python.
我想使用 python 复制 s3 存储桶中的文件。
Ex : I have bucket name = test. And in the bucket, I have 2 folders name "dump" & "input". Now I want to copy a file from local directory to S3 "dump" folder using python... Can anyone help me?
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“转储”和“输入”。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？
采纳答案by Felipe Garcia
Try this...
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)
[UPDATE] I am not a pythonist, so thanks for the heads up about the import statements. Also, I'd not recommend placing credentials inside your own source code. If you are running this inside AWS use IAM Credentials with Instance Profiles (http://docs.aws.amazon.com/IAM/latest/UserGuide/id_roles_use_switch-role-ec2_instance-profiles.html), and to keep the same behaviour in your Dev/Test environment, use something like Hologram from AdRoll (https://github.com/AdRoll/hologram)
[更新] 我不是 pythonist，所以感谢你对导入语句的提醒。另外，我不建议将凭据放在您自己的源代码中。如果您在 AWS 内部运行它，请使用 IAM 凭证和实例配置文件 ( 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 Hologram 之类的东西（https://github.com/AdRoll/hologram）
回答by Oren Efron
I used this and it is very simple to implement
我用过这个，实现起来很简单
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')
回答by vcarel
No need to make it that complicated:
没有必要让它变得那么复杂：
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)
回答by Manish Mehra
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)
回答by Piyush S. Wanare
This will also work:
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"
回答by Shakti
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)
回答by Boris
import boto3
s3 = boto3.resource('s3')
BUCKET = "test"
s3.Bucket(BUCKET).upload_file("your/local/file", "dump/file")
回答by Martin
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')
回答by Samuel Nde
This is a three liner. Just follow the instructions on the boto3 documentation.
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')
Some important arguments are:
一些重要的论据是：
Parameters:
参数：
str) -- The path to the file to upload.str) -- 要上传的文件的路径。str) -- The name of the bucket to upload to.
str) -- 要上传到的存储桶的名称。
str) -- The name of the that you want to assign to your file in your s3 bucket. This could be the same as the name of the file or a different name of your choice but the filetype should remain the same.Note:I assume that you have saved your credentials in a ~\.awsfolder as suggested in the best configuration practices in the boto3 documentation.
str) -- 您要分配给 s3 存储桶中的文件的名称。这可能与文件名或您选择的不同名称相同，但文件类型应保持不变。注意：我假设您已按照 boto3 文档中最佳配置实践的~\.aws建议将凭据保存在一个文件夹中。
回答by Willie Cheng
For upload folder example as following code and S3 folder picture
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: For more reference URL
PS：更多参考网址
