asked on

KORN Shell String Manipulation

The goal is to be able to strip out the first 3 octets of the ip addresses in /etc/hosts.

I can get the whole IP addresses with:

grep “\.” /etc/hosts | egrep –v “localhost|^#” | cut –d” “ –f1

...that yields something like:

10.209.2.45
10.209.6.24
:
:

but what I need to end up with is
10.209.2
10.209.6
:
:

Any help is most appreciated!

user_n

http://www.math.iitb.ac.in/resources/manuals/Unix_Unleashed/Vol_1/ch11.htm

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woolmilkporc

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dhite99

ASKER

user_n - that is a great reference, and sometime when I have several hours I'm going to look at it in detail. Regrettably today is not a day to do that, but thanks for the reference.

woolmilkporc - I like the idea of using awk, but I cannot seem to make it work:

oracle@emghlc001:/u01/app/oracle> awk -F\. '/^[0-9]{1,3}\./&&!/localhost/ {print $1 FS $2 FS $3}' /etc/hosts
oracle@emghlc001:>

It comes back with nothing.

/etc/hosts has something like this in it:

# Do not remove the following line, or various programs
# that require network functionality will fail.
127.0.0.1 localhost.localdomain localhost
::1 localhost6.localdomain6 localhost6
10.200.13.65 emrhlc001-MGMT
10.208.1.31 emrhlc001 emrhlc001.xxxx.com
10.208.1.34 emrhlc002 emrhlc002.xxxx.com
:
:

What am i doing wrong with the awk you provided?

dhite99

ASKER

I do not know what I would do without you guru's - thanks again!!!

Tintin

Here's some other methods

sed -n "/^[0-9]/ s/\.[0-9]* .*//p" /etc/hosts

grep "^[0-9]" /etc/hosts | cut -f1-3 -d.