Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
376 views
in Technique[技术] by (71.8m points)

python - Boto - Uploading file to a specific location on Amazon S3

This is the code I'm working from

import sys
import boto
import boto.s3

# AWS ACCESS DETAILS
AWS_ACCESS_KEY_ID = ''
AWS_SECRET_ACCESS_KEY = ''

bucket_name = AWS_ACCESS_KEY_ID.lower() + '-mah-bucket' conn = boto.connect_s3(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY)
bucket = conn.create_bucket(bucket_name, location=boto.s3.connection.Location.DEFAULT)
uploadfile = sys.argv[1]

print 'Uploading %s to Amazon S3 bucket %s' % 
       (uploadfile, bucket_name)

def percent_cb(complete, total):
    sys.stdout.write('.')
    sys.stdout.flush()

from boto.s3.key import Key
k = Key(bucket)
k.key = 'my test file'
k.set_contents_from_filename(testfile, cb=percent_cb, num_cb=10)

On my S3 I have created "directories", like this "bucket/images/holiday". I know these are only virtual directories.

My question is, how can I modify this upload specifically to bucket/images/holiday virtual directory on S3 rather than the bucket root?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

All you should have to do is prepend the virtual directory path to the key name prior to uploading. For example:

key_name = 'my test file'
path = 'images/holiday'
full_key_name = os.path.join(path, key_name)
k = bucket.new_key(full_key_name)
k.set_contents_from_filename(...)

You may have to change that a bit for your application but hopefully that gives you the basic idea.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...