small help in linux script sed command

dr.x · 04-11-2015, 05:04 PM

Hi Guys ,
i have a code that grap ipv6 address of the server
it seems like below :
ip addr show dev eth0 | sed -e's/^.*inet6 $[^ ]*$\/.*$/\1/;t;d'

2001:19f0:5c00:9422:5400:ff:fe07:b80d
fe80::5400:ff:fe07:b80d

as we see the result is 2 lines , but what i need is i need only the 1st line with only the 1st 4 octects
i mean i need only
2001:19f0:5c00:9422
the question is , what other shell command can i do so that i get only the 2001:19f0:5c00:9422 only ?

cheers

Didier Spaier · 04-11-2015, 06:41 PM

I would use a BRE for that:

Code:

bash-4.2# ip addr show dev eth1|grep dynamic|grep -o "\([[:alnum:]]*:\)\{4\}" 
2a01:e34:ec0b:630:
bash-4.2#

Caveat emptor: this solution is not fully portable, as the option -o is a GNU extension.

dr.x · 04-11-2015, 07:29 PM

Thnk you so much

syg00 · 04-11-2015, 09:28 PM

Similar regex works with sed - no need to shell out twice to grep

Code:

sed -nr '/inet6/ s/.*[[:space:]](([[:alnum:]]*:){4}).*/\1/p'