"I want to extract everything contained within the first pair of parentheses in the given string"

In that case....

#!/usr/bin/perl -w
use strict;

my $string = "Identities = 1529/1540 (99%), Gaps = 2/1540 (0%)";
my ($wanted) = $string =~ m#\((.*?)\)#;
print "$wanted\n";
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what does the m# do? m is for match, but I don't understand what the # does.

Hi rangersfan,

(/) This is the usual delimiter for the text part of a regular expression. you can also use (#) as delimiter. By using usual delimiter with regexp i have given the code below you can also refer that for your understand :-p

You haven't said what's variable and what's fixed so there are any number of possible solutions. If you want to remove the 99%, you can just say $string =~ s/99%//; or if you want to remove all things that look like percentages, you could say $string =~ s/\d+%//g;. But you may want something more complicated. I don't know.

Hi try this,

#!/usr/local/bin/perl -w
use strict;
my $str = "Identities = 1529/1540 (99%), Gaps = 2/1540 (0%)";
my ($output) = $str =~ m/(\d+%)/gsi;
print "$output\n";
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use strict;
use warnings;

my ($text, %percent);

$text = 'Identities = 1529/1540 (99%), Gaps = 2/1540 (0%)';
$percent{$1} = $2 while $text =~ /(\w+) = \d+\/\d+ \((\d+)%\)/g;

for (sort keys %percent) {
    print "$_ : $percent{$_}\n";
}
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