in reply to reg ex help

If you know the separator (white space) you don't have to use a regexp. A simple split will do. 
my $line = qq|30 44 02 01 01 04 06 70 75 62 6C 69 63 A7 37 02 0D.....p
+ublic.7|;
my $fourteenthnumber = (split (/ /, $line))[13];
print $fourteenthnumber;

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updated with tachyon's syntactic sugar from /\s/ to / /.

pelagic

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Re: Re: reg ex help
by tachyon (Chancellor) on Apr 15, 2004 at 10:19 UTC

    FWIW split ' ', $data is magical in that the ' ' matches any quanitity of whitespace ie spaces, tabs, newlines. Syntactic sugar for \s+ really.

    cheers

    tachyon

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      FWIW split ' ', $data is magical in that the ' ' matches any quanitity of whitespace ie spaces, tabs, newlines. Syntactic sugar for \s+ really.
      split ' ' and split /\s+/ are not quite the same: 
      #!/usr/bin/perl

use strict;
use warnings;

$_ = " foo bar baz ";

my @a = split ' ';
my @b = split /\s+/;

print scalar @a, "\n";
print scalar @b, "\n";
 
__END__
3
4

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      Abigail

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        Thanks for clarifying that ' ' is DWIM (magical) with respect to leading whitespace. 

        #!/usr/bin/perl

use Data::Dumper;

$_ = " foo bar baz ";

my @a = split ' ';
my @b = split /\s+/;

print Dumper \@a;
print Dumper \@b;

__DATA__
$VAR1 = [
          'foo',
          'bar',
          'baz'
        ];
$VAR1 = [
          '',
          'foo',
          'bar',
          'baz'
        ];

        [download]

        What interests me from a 'why is it so?' point of view is that if \s+ splits at the begining of a string to find the NULL why not at the END as well. Your example documants the behaviour but why is leading whitespace treated differently from trailing whitespace? There is afterall a null string after the trailing whitespace as well.

        cheers

        tachyon

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