[SOLVED] awk expression that works on awk v4.0.2 but it does not on >= 4.2.1

Issue

I have this awk command:

echo www.host.com |awk -F. '{$1="";OFS="." ; print $0}' | sed 's/^.//'

which what it does is to get the domain from the hostname:

host.com

that command works on CentOS 7 (awk v 4.0.2), but it does not work on ubuntu 19.04 (awk 4.2.1) nor alpine (gawk 5.0.1), the output is:

host com

How could I fix that awk expression so it works in recent awk versions ?

Solution

For your provided samples could you please try following. This will try to match regex from very first . to till last of the line and then prints after first dot to till last of line.

echo www.host.com | awk 'match($0,/\..*/){print substr($0,RSTART+1,RLENGTH-1)}'

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OP's code fix: In case OP wants to use his/her own tried code then following may help. There are 2 points here: 1st- We need not to use any other command along with awk to processing. 2nd- We need to set values of FS and OFS in BEGIN section which you are doing in everyline.

echo www.host.com | awk 'BEGIN{FS=OFS="."} {$1="";sub(/\./,"");print}'

Answered By - RavinderSingh13

Answer Checked By - David Goodson (WPSolving Volunteer)