Select an IP address in a regex group from a multi line string

Question

I'm trying to select an IP address from a routing table and am not having much luck crafting a sed or grep expression.

I have the below regex that is valid and selects group 1 as expected in regex testers.

.*0\.0\.0\.0\/0.*(\b\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\b).*eth0.* 

I've been able to easily get the output I need via awk, however the bash variable is null when the script is executed in my environment. I've tried various expressions with sed and awk, but there seems to be more selectors required than I'm familiar with

/usr/bin/vtysh -c 'show ip route' | sed -n 's/.*\(\b\d\{1,3\}\.\d\{1,3\}\.\d\{1,3\}\.\d{1,3}\b\).*eth0.*/\1/p' 

The input is below

Codes: K - kernel route, C - connected, S - static, R - RIP, O - OSPF, I - IS-IS, B - BGP, E - EIGRP, N - NHRP, T - Table, v - VNC, V - VNC-Direct, A - Babel, D - SHARP, F - PBR, f - OpenFabric, > - selected route, * - FIB route, q - queued route, r - rejected route S>* 0.0.0.0/0 [210/0] via 9.16.24.1, eth0, 1d06h57m S> 0.0.0.0/0 [300/0] via 9.16.26.1, eth1, 1d06h57m O>* 10.0.0.0/19 [110/42] via 10.255.255.21, eth1, 1d06h56m * via 10.255.255.22, eth1, 1d06h56m 

expected output:

9.16.24.1 

Answer 1

I think this is a job for grep (GNU) or awk, rather than sed:

grep -Po "[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}(?=, eth0)" file 

Ouput:

9.16.24.1