Two Amazon S3 Python Boto Upload Questions
amazon-s3, boto, python
Solution
Actually, instead of conn.create_bucket(...), just do:
bucket = conn.get_bucket(bucket_name)
k = Key(bucket)
Also, I should probably point out that it seems silly to me to name your buckets based on the AWS_ACCESS_KEY_ID, as this is not really needed.
Problem
I am working with this code currently: ``` #!/usr/bin/env python import boto import boto.s3 from boto.s3.key import Key AWS_ACCESS_KEY_ID = '' AWS_SECRET_ACCESS_KEY = '' filename = 'test.zip' 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) k = Key(bucket) k.key = 'my test file' k.set_contents_from_filename(filename) ``` I have two questions. Firstly, I believe this code creates a bucket as well as doing an upload. Thing is I don't want a bucket created, since I already have one in place. In order to do this do I just change it to this: ``` k = Key(bucket_name) ``` and get rid of this: ``` bucket = conn.create_bucket(bucket_name, location=boto.s3.connection.Location.DEFAULT) ```