How regex works here?

nagalenoj has asked for the wisdom of the Perl Monks concerning the following question:

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";
[download]

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/;

Comment on How regex works here? Select or Download Code

Replies are listed 'Best First'.
Re: How regex works here? by lakshmananindia (Chaplain) on Apr 08, 2009 at 05:27 UTC
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 $lines has only the matched pattern. --Lakshmanan G. The great pleasure in my life is doing what people say you cannot do.	[reply]
Re^2: How regex works here? by nagalenoj (Friar) on Apr 08, 2009 at 05:38 UTC
yeeeah..! Fine.. I understood., So the problem is not with the matching and with the substitution. Thanks..!	[reply]
Re^3: How regex works here? by bart (Canon) on Apr 08, 2009 at 09:01 UTC
Yes. There's no need at all to change the original string to extract the data you want. All you need is to match it. `if($line =~ /^(\S+\s+\S+)/) { $match = $1; }` [download] or even simpler: `($match) = $line =~ /^(\S+\s+\S+)/;` [download] 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.	[reply] [d/l] [select]
Re: How regex works here? by codeacrobat (Chaplain) on Apr 08, 2009 at 06:36 UTC
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";` [download] `print+qq(\L@{[ref\&@]}@{['@'x7^'!#2/"!4']});`	[reply] [d/l] [select]
Re: How regex works here? by AnomalousMonk (Archbishop) on Apr 08, 2009 at 07:11 UTC
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'` [download]	[reply] [d/l]