Re: Re: Multiple matches of a regex

Thanks for the tip, Fletch! (Can't ++ you today, ran out of votes... Wait for monday please!) So that leaves me with:

use strict;

my $regex = qr/(\d\d):(\d\d)(?::(\d\d))?/;

#             01234567890123456
my $string = "11:30 or 11:29:53";

my $res;
while ($string =~ /$regex/g) {
    my $pos = $-[0];
    my ($ind, @match);
     for $ind (1..$#-) {
         push @match, substr ($string, $-[$ind], $+[$ind] - $-[$ind]);
    }
    $res->{$pos} = \@match;
}
use Data::Dumper;
print Data::Dumper::Dumper $res;
[download]

This prints:

as it should.

What I don't like about this solution is that it involves a lot of substr calls to get strings that are stored in the $n variables anyway. Is there a better way to do this?

Thanks,

pike

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Re: Re: Re: Multiple matches of a regex by BrowserUk (Patriarch) on Jan 10, 2003 at 19:09 UTC
As you are using capturing brackets within your regex, your required substrings are already being placed into the variables $1 $2 $3, so there is no need to use substr to isolate the substrings a second time. The only tricky bit is that $3 may have no value, so I've used grep to exclude it if it has no value. `#! perl -slw use strict; use Data::Dumper; my $regex = qr/(\d\d):(\d\d)(?::(\d\d))?/; # 01234567890123456 my $string = "11:30 or 11:29:53"; my $res; $res->{$-[0]} = [grep $_, $1, $2, $3] while $string =~ /$regex/g; print Dumper $res; __END__ c:\test>225814 $VAR1 = { '0' => [ '11', '30' ], '9' => [ '11', '29', '53' ] }; c:\test>` [download] Examine what is said, not who speaks. The 7th Rule of perl club is -- pearl clubs are easily damaged. Use a diamond club instead.	[reply] [d/l]
Re: Re: Re: Multiple matches of a regex by ihb (Deacon) on Jan 10, 2003 at 19:09 UTC
If you have anything against `substr` calls you can do this symbolically (through more people will probably have something against this than repeated `substr` calls). `no strict 'refs'; my @groups = map $$_, 1 .. $#+;` [download] If I shall be picky, in your question you wanted the first match to return `['11', '30', undef]` while in your reply you said `['11', '30']` is what you want. As I pointed out here you should consider the difference between `$#+` and `$#-`. Also, unmatched groups would be `''` with your code. Perhaps that's what you want, but it won't be fully analogous to using the `$<DIGITS>` variables. If done symbolically it sure will be analogous though, since you actually use the `$<DIGITS>` variables. Hope I've helped, `ihb`	[reply] [d/l] [select]