Issue
I would like to filter the first IP from the given output below.
3676798 I/H 628 71.1.219.106 10198 71.1.208.58 20198 UDP A
3676798 R/U 628 71.1.208.58 20198 71.1.219.106 10198 UDP A
3676799 I/H 1066 71.1.57.57 10008 71.1.57.53 20008 UDP A
3676799 R/U 1066 71.1.57.53 20008 71.1.57.57 10008 UDP A
3676800 I/H 532 71.1.213.104 10142 71.1.203.52 20142 UDP A
3676800 R/U 532 71.1.203.52 20142 71.1.213.104 10142 UDP A
Using the below command, I'm able to get all the IP's but I would like to get only the first ip from each line which has 2 IP's. There may be line which might not have IP's at all. so I would like to ignore that line.
grep -o '[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}' <filename>
I tried to use Awk to get it (as suggested in one of the comments) but this was unsuccessful since it prints the fourth field from any line which does not have IP's as well.
awk '{ print $4 }' file
Any pointers would be highly helpful.
Solution
To only print when the fourth field matches an IP address,
awk '$4 ~ /^[1-9][0-9]{1,2}(\.[0-9]{1,3}){3}$/{ print $4 }' file
To only print the first IP address on any line, regardless of where exactly it's found, try Perl:
perl -ne 's/.*?([0-9]{1,3}(\.[0-9]{1,3}){3}).*/$1/ and print' file
This looks quite similar to sed but the non-greedy .*? is not available in sed. If you wanted the last IP address on every line, it would be easy to do with sed.
Answered By - tripleee Answer Checked By - Willingham (WPSolving Volunteer)