Finding all Amazon AWS Instances That Do Not Have a Certain Tag

Question

I\'m trying to use the Amazon AWS Command Line Tools to find all instances that do not have a specified tag.

Finding all instances WITH a tag is simple enough, e.g.<

Answer 1

I too was totally shocked by how difficult this is to do via the CLI. I liked user2616321's answer, but I was having a little trouble making it output the exact fields I wanted per instance. After spending a while messing around and failing with JMESPath in the query syntax, I ended up just making a little ruby script to do this. In case anyone wants to save a few minutes writing one of their own, here it is:

#!/usr/bin/env ruby
require 'json'

# We'll output any instance that doesn't contain all of these tags
desired_tags = if ARGV.empty?
                 %w(Name)
               else
                 ARGV
               end

# Put the keys we want to output per instance/reservation here
reservation_keys = %w(OwnerId RequesterId)
instance_keys = %w(Tags InstanceId InstanceType PublicDnsName LaunchTime PrivateIpAddress KeyName) 
instances_without_tags = []

# Just use CLI here to avoid AWS dependencies
reservations = JSON.parse(
  `aws ec2 describe-instances`
)["Reservations"]

# A reservation is a single call to spin up instances. You could potentially
# have more than one instance in a reservation, but often only one is
# spun up at a time, meaning there is a single instance per reservation.
reservations.each do |reservation|
  reservation["Instances"].each do |instance|
    # Filter instances without the desired tags
    tag_keys = instance["Tags"].map { |t| t["Key"] }
    unless (tag_keys & desired_tags).length == desired_tags.length
      instances_without_tags << 
        reservation.select { |k| reservation_keys.include?(k) }.
          merge(instance.select { |k| instance_keys.include?(k) })
    end
  end
end

puts JSON.pretty_generate(instances_without_tags)