Bashscripts.org

Hello,

I am reaching out to the pro's here. In one of my network bash scripts, I need to be able to print only the first three octets of on IP address. Here is what I have so far:

This would trim off the last octet for 192.168.10.100, giving me:
192.168.10.

but I need this to be universal enough to handle any class of IP address...say a class A IP of 10.100.15.10? This line would give me:
10.100.15.1

I still need that last "1" trimmed. I am sure there is a better way to do this, so I don't have to modify the script itself each and every time I need to do this task.

btw...if anyone was wondering why I am removing the loopback, or why it is even coming up at all...this is the only way I could figure out how to get just the static IP from the ifconfig. If there a better way for that, then I am all ears (ps...I tried using hostname -i, but when this script runs, the hosts file isn't set up yet and this defaults to the loopback).

Any help is most appreciated!

well, I think I solved it:


this will give me the first three octets (only missing the final ".", but I can live with that and put that into where I echo the variable.

But, I am sure I wrote this string way too long, so if anyone has a way to shorten it and get the same results, I would love to learn.

don't pipe grep to awk, and then to sed, or/and cut. If you think about such an awful thing, remember that awk can do the job alone.

Practically the same as Watael, but slightly shorter:

or

or

...