Perl - substring keywords - perl

Question

I have a text file where is lot of lines, I need search in this file keywords and if exist write to log file line where is keywords and line one line below and one above the keyword. Now search or write keyword not function if find write all and I dont known how can I write line below and above. Thanks for some advice.
my $vstup = "C:/Users/Omega/Documents/Kontroly/testkontroly/kontroly20220513_154743.txt";
my $log = "C:/Users/Omega/Documents/Kontroly/testkontroly/kontroly.log";
open( my $default_fh, "<", $vstup ) or die $!;
open( my $main_fh, ">", $log ) or die $!;
my $var = 0;
while ( <$default_fh> ) {
if (/\Volat\b/)
$var = 1;
}
if ( $var )
print $main_fh $_;
}
}
close $default_fh;
close $main_fh;

Answer 1

The approach below use one semaphore variable and a buffer variable to enable the desired behavior.
Notice that the pattern used was replaced by 'A` for simplicity testing.
#!/usr/bin/perl
use strict;
use warnings;
my ($in_fh, $out_fh);
my ($in, $out);
$in = 'input.txt';
$out = 'output.txt';
open($in_fh, "< ", $in) || die $!."\n";
open($out_fh, "> ", $out) || die $!;
my $p_next = 0;
my $p_line;
while (my $line = <$in_fh>) {
# print line after occurrence
print $out_fh $line if ($p_next);
if ($line =~ /A/) {
if (defined($p_line)) {
# print previous line
print $out_fh $p_line;
# once printed undefine variable to avoid printing it again in the next loop
undef($p_line);
}
# Print current line if not already printed as the line follows a pattern
print $out_fh $line if (!$p_next);
# toggle semaphore to print the next line
$p_next = 1;
} else {
# pattern not found.
# if pattern was not detected in both current and previous line.
$p_line = $line if (!$p_next);
$p_next = 0;
}
}
close($in_fh);
close($out_fh);