in reply to Regex match on implicit default variable ($_) in a script, not a one liner

I want to remove the middle line from $A. Same as I did with the one liner.
What is desired:
Before: $A="abc\ndef\nghi\n"
After: $B="abc\nghi\n"

Heres another way, which you can also iterate through the array easily (once you have built one ofcourse) :) 

use strict;
use warnings;

my @array = "abc\ndef\nghi\n";
print $1 . $3 if $array[0] =~ /(.*)(\ndef)(.*)/s;

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Re^2: Regex match on implicit default variable ($_) in a script, not a one liner
by Laurent_R (Canon) on Oct 24, 2015 at 18:43 UTC
    I do not see your point about using an array here, since you are not really using an array but just one element of your array, so that a scalar value would just work the same way: 
    my $scalar =   "abc\ndef\nghi\n";
print $1 . $3 if $scalar =~ /(.*)(\ndef)(.*)/s;

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[d/l]

      I was trying to duplicate what the one liner does, so I wanted to keep it as a string.
      I finally figured it out: preserve $1 and use "if" instead of "while":

      #!/usr/bin/perl

# full script the following one liner:
# $ echo -e "abc\ndef\nghi\n" | perl -wlne '! /def/ and print "$_";'
# abc
# ghi

use strict;
use warnings;
use diagnostics;

my $C;
my $B = "";
my $A = "abcdef
fhijkl
mnopqr
";

while ( $A =~ m{(.+)}g ) {
$B .= "$1\n" unless $1 =~ /ijk/;
# $B .= "$1\n" unless $1 =~ /xyz/;
}
print "\$B = \n$B\n";


$B = "";
while ( $A =~ m{(.+)}g ) {
$C = $1;
$B .= "$C\n" if $C =~ /ijk/;
}
print "\n\$B = \n$B\n";

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      $B =
      abcdef
      mnopqr


      $B =
      fhijkl

      Thank you for all the help! -T

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