A shell script replacement for the Dropbox public folder, using S3 for storage and serving.

share.bash

#!/usr/bin/env bash

#
# Usage:
#
#    share.bash file1 file2 file3 ...
#

#
# Simple setup:
#
# 1. Install the AWS CLI tool:
#    http://docs.aws.amazon.com/cli/latest/userguide/installing.html
#
# 2. Sign up for an Amazon Web Services account at https://aws.amazon.com/ 
#    (don't forget to set up billing to prevent service interruptions)
#
# 3. Navigate to Amazon S3 on the AWS website.
#
# 4. Create an S3 bucket; give it any name you like, create it in any region
#    you like, and copy the name into the "S3_BUCKET" setting into this file.
#
# 5. In the top-right of the AWS website, click your name and open 
#    "My Security Credentials"
#
# 6. Open "Access Keys (Access Key ID and Secret Access Key)"
#
# 7. Click "Create New Access Key"
#    IMPORTANT: This access key allows FULL ACCESS to your account! Keep it
#               secret! Don't share this file with ANYONE after you've entered
#               it!
#               A better option is to create an IAM user with access only to S3
#               but that's beyond the scope of these "Simple setup" instructions.
#
# 8. Copy the new Access Key ID and Secret Access Key into this file.
#

##
## Settings
##

# Enter your S3 bucket name
export S3_BUCKET='mysweetbucket'

# Enter your Amazon access key ID
export AWS_ACCESS_KEY_ID='AKIA...etc'

# Enter your Amazon secret access key
export AWS_SECRET_ACCESS_KEY='aAbuhoJFWNGK/etc'

##
## Don't change anything after this point!
##

if [[ $# == 0 ]]
then
    echo "Usage: $0 file1 file2 file3 ..."
    exit 1
fi

BUCKET_REGION="$(aws s3api get-bucket-location --bucket "$S3_BUCKET" --query 'LocationConstraint' --output text)"
export AWS_DEFAULT_REGION="$BUCKET_REGION"

i=0
for filename in "$@"
do
    if [[ $i > 0 ]]; then
        echo
    fi

    echo "upload $filename"
    aws s3 cp --storage-class STANDARD_IA --acl public-read "$filename" s3://"$S3_BUCKET"/
    if [[ $? == 0 ]]; then
        urlescaped="$(python -c 'import urllib, sys; print(urllib.quote(sys.argv[1]))' "$filename")"
        echo "https://s3.${BUCKET_REGION}.amazonaws.com/${S3_BUCKET}/$urlescaped"
    fi

    i=$((i + 1))
done