Find available IP addresses using DNS and DIG ( -x axfr )

available_ips.sh

#!/usr/bin/env bash

# Find available IP addresses from by using dig -x axfr 

usage() {
  cat <<-EOF
  usage: ${0} <network>
  example: ${0} 10.1.10
EOF
}

declare -r network=$1

if [ -z "${network}" ]; then
  echo "ERROR: No network provided"
  usage
  exit 1
fi

# Reference Commands 
cmd_list='dig sort awk'

# Function to check if referenced command exists
cmd_exists() {
  if [ $# -eq 0 ]; then
    echo 'No command argument was passed to verify exists'
    exit 1
  fi

  cmd=${1}
  cmd_fullpath=$(which "${cmd}")

  if [ ! -x "${cmd_fullpath}" ];  then
    echo "${cmd} command not found"
    exit 1
  #else
  #  echo "Found: ${cmd} at ${cmd_fullpath}"
  fi
}

# Verify that referenced commands exist on the system
for cmd in ${cmd_list}; do
  cmd_exists "$cmd"
done

# Reference: http://stackoverflow.com/questions/15867557/finding-gaps-sequential-numbers
# Explanation of [ awk '$1!=p+1{print p+1"-"$1-1}{p=$1}' ]
#  *  $1 is the first column from current input line
#  *  p is the previous value of the last line
#  *  so ($1!=p+1) is a condition : if $1 is different than previous value +1, then :
#  *  this part is executed : {print p+1 "-" $1-1} : print previous value +1, the - character and fist columns + 1
#  *  {p=$1} is executed for each lines : p is assigned to the current 1st column

available_ips=$(dig -x "${network}" axfr | sort -k 1 -n -u | awk -F'.' '/^[0-9]/ { print $1 }' | awk '$1!=p+1{print p+1"-"$1-1}{p=$1}')

for i in ${available_ips}; do
  seq1=$(echo "${i}" | cut -d'-' -f1)
  seq2=$(echo "${i}" | cut -d'-' -f2)

  if [ "${seq1}" == "${seq2}" ]; then
    echo "${network}.${seq1}"
  else
    echo "${network}.${i}"
  fi
done