awk help

[srikanthradix]

server 12:10 uid:gid> ls -l app-validator
lrwxrwxrwx 1 skedev etdit 24 Aug 24 06:52 app-validator -> app-validator-1.3.22/


server 12:10 uid:gid> ls -l app-validator | awk '{print $11}' | awk -F 'app-validator-' '{print $2}' | awk -F '/' '{print $1}'

1.3.22

I want to get 1.3.22. Is there anything that is more terse and effective than the above awk statement? Please suggest.

[tpl]

welcome to the forum

you could use "cut" twice--

cut -d - -f5 | cut -d / -f1

[srikanthradix]

Thanks for the reply. It works perfectly.