Regex to match IP addresses but ignore localhost

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So I have this script that does something with IPs allocated to my OS (GNU/Linux) that I get from running ifconfig. It works fine, however, I was wondering if I could filter out loopback/localhost IP (127.0.0.1) in the same regex expression [I assume every server within my cluster has said IP and I don't need to do anything with it in my script.]

What my script uses is: 

ifconfig | awk '/(([0-9]{1,3}\.){3})/ {print}' |sed -e "s/.*addr\://g" -e "s/\s.*//g"

I get results like:

> ifconfig | awk '/(([0-9]{1,3}\.){3})/ {print}' |sed -e "s/.*addr\://g" -e "s/\s.*//g"
172.16.0.1
127.0.0.1

I know it might be a stupid question, but could I filter any IP that starts with 127 in my first regex?

I could try changing awk for grep, somethin like:

> ifconfig |egrep -o "addr\:(([0-9]{1,3}\.){3}[0-9]{1,3})" |sed -e "s/.*addr\://g"

but if I try to negate (?!127) at the beginning, bash will interpret it as !127 which would just throw me something from the history.

I mean, I could just run another grep at the end of the oneliner like grep -v "127.0.0.1", but I just wanted to avoid greping something already greped. Not that anything is wrong with that, just trying to know little more and be more efficient, I guess.

regex-negation grep awk linux bash
2 Answers

With only one grep without sed or awk:

# ip a|grep -oP "inet \K[0-9.]*(?=.*[^ ][^l][^o]$)"
192.168.1.31
172.16.5.31
user_64811210

You can just add a clause to match the 127.0.0.1 and exclude it by adding the next as below. This way Awk ignores doing any action on the lines containing this pattern.

.. | awk '/127.0.0.1/{next}/(([0-9]{1,3}\.){3})/{print}' | ..
user_52910150
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