How regex works here?

[nagalenoj]

Dear monks,

I have to take the first 2 columns out from a line(using regular expression). For that I have written a code like follows,

use strict;
use warnings;

$lines = 'B6y green $1$Byf3501d$dPtaW2vCbq63sdsw';

$lines =~ s/^(\S+\s\S+)/$1/;
print "$lines";

But, its matching the whole line. why?

When I have changed the regex as follows, its working fine. How both differs. Why the first one doesn't work?

$lines =~ s/^(\S+\s\S+).*/$1/;

[lakshmananindia]

In the first regex you are matching only the "B6y green" and you are replacing that again with "B6y green". So the $lines is same as it is

In the second regex you are matching "B6y green" and the remaining string and you are replacing that with "B6y green". So the $lines has only the matched pattern.

[nagalenoj]

yeeeah..! Fine.. I understood.,

So the problem is not with the matching and with the substitution.

Thanks..!

[bart]

Yes.

There's no need at all to change the original string to extract the data you want. All you need is to match it.

if($line =~ /^(\S+\s+\S+)/) {
    $match = $1;
}

or even simpler:

($match) = $line =~ /^(\S+\s+\S+)/;

Note that for the latter, the assignment must be in list context, not scalar context; that's what the parens around the variable on the left hand side are for.

[codeacrobat]

have a look :-)

use strict;
use warnings;
use re qw(debugcolor);
my $lines = 'B6y green $1$Byf3501d$dPtaW2vCbq63sdsw';

$lines =~ s/^(\S+\s\S+)/$1/;
print "$lines";

[AnomalousMonk]

In contrast, consider matching and deleting only that which you wish to eliminate from the string:

>perl -wMstrict -le
"my $line = 'B6y green $1$Byf3501d$dPtaW2vCbq63sdsw';
 $line =~ s{ \s \S+ $ }{}xms;
 print qq{'$line'};
"
'B6y green'