Boto – Uploading file to a specific location on Amazon S3

Question

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?

Asked By: Jimmy

||

Source

Answer 1

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.

Answered By: garnaat

Answer 2

You can also use this:

from boto.s3.key import Key

bucket = conn.get_bucket('images')
k = Key(bucket)
k.key = 'holiday/my_test_file.txt'

It works like this in my s3_percent_uploader

python.exe s3_percent_uploader.py test_upload.txt testbucket test/upload.txt

after upload file will be in testbucket/test/upload.txt

Answered By: olekb

Boto – Uploading file to a specific location on Amazon S3

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