Question

Is there a more efficient way to get this done? By this I mean can I refactor my script to get the outcome more efficiently, not getting this info from another source ie. Amazon via API or SDK.

I do a curl that returns this

{
         "instanceId" : "i-123abcde",
         "billingProducts" : null,
         "version" : "2001-01-01",
         "accountId" : "123456789012",
         "instanceType" : "m1.large",
         "architecture" : "x86_64",
         "kernelId" : "aki-123ab1ab",
         "ramdiskId" : null,
         "pendingTime" : "2001-01-00T00:00:00Z",
         "availabilityZone" : "us-east-1a",
         "imageId" : "ami-12345ab1",
         "devpayProductCodes" : null,
         "privateIp" : "10.10.10.10",
         "region" : "us-east-1"
         }

This is my code block

require 'rubygems'
require 'curb'
require 'pp'

info = Curl.get("http://169.254.169.254/latest/dynamic/instance-identity/document")

foo = info.body_str.split(',').join.scan(/"(.*?)" : "(.*?)"/)

quux = {}
foo.each do |baz|
        quux[baz[0]] = baz[1]
end
pp quux

The outcome is a hash with keys

{
         "instanceId" => "i-123abcde",
         "version" => "2001-01-01",
         "accountId" => "123456789012",
         "instanceType" => "m1.large",
         "architecture" => "x86_64",
         "kernelId" => "aki-123ab1ab",
         "pendingTime" => "2001-01-00T00:00:00Z",
         "availabilityZone" => "us-east-1a",
         "imageId" => "ami-12345ab1",
         "privateIp" => "10.10.10.10",
         "region" => "us-east-1"
         }

Answer 1

That is JSON data. Do this:

require 'json'
...
quux = JSON.parse(info.body_str)