[SOLVED] awk question - FS/RS works in 1 instance, doesn't in another

psycroptic · 08-27-2013, 09:02 PM

i am woefully unfamiliar with awk. i am wanting to return whatever is in between 2 specific words in a string. I have this:

Code:

ip addr show external | grep inet

which gives me this:

Code:

    inet xx.xx.xx.xx/22 brd 255.255.255.255 scope global external

I want to return that address between "inet" and "brd". So i do this:

Code:

ip addr show external | grep inet | awk 'BEGIN{FS="inet";RS="brd"}/</{print $2}'

Which i would think should work, as I've used that RS= FS= thing before with nearly-identical syntax and expected behavior, but here I get no output at all.

Ideas?

Firerat · 08-27-2013, 10:24 PM

looks like you are over thinking it..

Code:

ip addr show external | awk '/inet/{print $2}'

https://www.gnu.org/software/gawk/manual/

Edit:

actually, if you have inet6 then..

Code:

awk '$1 == "inet" {print $2}'
# or
awk '/inet /{print $2}'

grail · 08-28-2013, 08:53 AM

Personally I would have never expected the awk example to work as the regex you are using does not exist, ie. '<' is no where in the line shown

psycroptic · 08-28-2013, 10:10 AM

cool thanks, the thing firerat posted worked. as i said, i'm not too familiar with awk.

aprreciate it