Bash comprehensive list of IP addresses for a domain

Question

I'm trying to produce a list of all the possible ip addresses for a given domain name. I think I'm close but don't know what I'm missing (or if there is a better way).

First I create a list of variations of the domain like so:

 webkinz.com
 www.webkinz.com

I then loop over this list and run dig on each variation like so:

 while read domain; do
    IPs=`dig $domain | grep $domain | grep -v ';' | awk '{ print $5 }'`;
    echo " ${IPs}" >> /tmp/IPs; #array
 done < /tmp/mylist

 sort -u /tmp/IPs > /tmp/TheIPs; #remove duplicates
 cat /tmp/TheIPs| tr -d "\n" > /tmp/IPs  #remove new lines (making it 1 long line)

My IPs file looks like this:

  66.48.69.100 www.webkinz.com.edgesuite.net.a1339.g.akamai.net.

Only 3 problems. :-(

1. Dig returned domains when I was only expecting ip addresses.
2. Some how my script deleted the spaces between the domains.
3. Some of the ip addresses from dig www.webkinz.com are missing.

So, how should I do this? Do I somehow figure out if dig returned another domain instead of an ip address and run dig on that domain? Do I just ignore domain names returned from dig and figure the ip addresses is sufficient? I want to catch every ip address that will resolve to the domain if possible. I didn't think it should be this hard. Any ideas?

Answer 1

In order to get just the IP addresses, use dig +short:

#!/bin/bash
while read -r domain
do
    dig +short "$domain"
done < /tmp/mylist | sort -u | awk '{printf "%s ", $0} END {printf "\n"}' > outputfile

or

#!/bin/bash
echo $(xargs -a /tmp/mylist dig +short | sort -u) > outputfile

Using echo with an unquoted argument drops the newlines except at the end.

You don't need any intermediate variables or temporary files.