Argthtjtr

I've got a screw I'm trying to nail in with perl and so far this is what I've got so far.

perl -ne '/(.+).(.+)((.+))(.+)(?{print "match" if  ( $1 > 9 || ( $1 == 9 && $2 > 1 ) || ($1 == 9 && $2 == 1 && $3 > 7 ) || $1 == 9 && $2 == 1 && $3 == 7 && $4 > 22 ) })/' versions

versions:

9.1(7)23

9.1(7)22

8.1(7)22

7.2(33)24

55

it will print "match" if the version in the file is > than 9.1(7)23, which is good.

But I want the regexp to succeed, not just print "match". How do I translate the stuff inside the code block to an actual response. I've tried quite a few iterations with *ACCEPT|*FAIL but nothing has worked so far.

First of all, the regex does succeed. But it does so whether the condition is true or not. I think you're actually asking for it to fail when the condition is false. For that, you want

(?(?{ condition })(*ACCEPT)|(*FAIL))

or just

(?(?{ !condition })(*FAIL))

Fixed:

perl -nle'

   print "$_: match"

      if /

         ^(.+).(.+)((.+))(.+)z

         (?(?{

            !( $1 > 9

            || $1 == 9 && $2 > 1

            || $1 == 9 && $2 == 1 && $3 > 7

            || $1 == 9 && $2 == 1 && $3 == 7 && $4 > 22

            )

         })(*FAIL))

      /x;

' versions

A far better approach is to do the check outside of the pattern.

perl -nle'

   print "$_: match"

      if /^(.+).(.+)((.+))(.+)z/

      && (

            $1 > 9

         || $1 == 9 && $2 > 1

         || $1 == 9 && $2 == 1 && $3 > 7

         || $1 == 9 && $2 == 1 && $3 == 7 && $4 > 22

      );

' versions