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

Fellow monks,

I know I'm probably going to slap my forehead after I get a response for this but...

I've got a very simple regex which I want to just pull sequential digits out of a filename which is passed in - e.g. sfytd13.txt should return 13.

I'm using the following code: 

use strict;

my $file = 'file013.txt';
my $sk = $file =~ m/\d+/;

print $sk;

[download]

Very simple, but the problem is that I'm only getting the '1' back. I realize that if I use parthensis matching $1 will return '013' but I'm curious as to why my code doesn't return 13.

My understanding is that \d+ is going to match a digit followed by any number of digits until I hit a non-digit. What am I missing here? Is something getting clobbered in the assignment?

Thanks in advance!

There is no emoticon for what I'm feeling now.

Replies are listed 'Best First'.
Re: Regex not behaving as expected
by Enlil (Parson) on Feb 01, 2003 at 00:38 UTC
    this works (assuming you want a series of numbers but disregarding leading zeros): 
    use strict;

my $file = 'file013.txt';
my ($sk) = $file =~ m/0*(\d+)/;

print $sk;

    [download]
    The problem with the way you had it is that it is returning 1 not because of the number 1 in the 013 but because it is true (akai.e., there was a match).

    -enlil

[reply]
[d/l]
      Yes that's exactly what I was after. Thanks for that, I think it's been too long of a day. I can't believe I didn't catch that. :|

      Also thank you for the ($sk) bit. I never realized that you didn't have to use $1, $2, etc... in matching.

      There is no emoticon for what I'm feeling now.

[reply]
Re: Regex not behaving as expected
by Paladin (Vicar) on Feb 01, 2003 at 00:33 UTC
    m// in scalar context returns true or false if it succeeds or fails. In list context without /g it returns the list of $1, $2, ... if you have used () to capture anything. In short you probably want to use: 
    use strict;

my $file = 'file013.txt';
my ($sk) = $file =~ m/(\d+)/; # list context to get what () captured

print $sk;

    [download]
    See perldoc perlop for more info on various combinations of m//, /g and what they return in list and scalar context.

    Update: Corrected typo.

[reply]
[d/l]
[select]
(elbie): Regex not behaving as expected
by elbie (Curate) on Feb 01, 2003 at 15:44 UTC

    Your solution looks very close to a common trick I use, which is to assign a vairable to another variable, then do a regex search and replace on the new one: 

    my $file = 'file013.txt';
(my $sk = $file) =~ s/^.*?(\d+).*$/$1/;

    [download]

    Which for the sake of this problem is a very similar solution to what others are proposing, the above method is much more useful when I want to keep most of the string rather than a portion of it.

    Example, I want to emphasise the word keep above: 

    my $boring_string = '...when I want to keep most of the string...';
(my $exciting_string = $boring_string) =~ s/(keep)/<b>$1</b>/;

    [download]

    elbieelbieelbie

[reply]
[d/l]
[select]

      If you use .* as "everything else", then be sure to use the /s modifier.   s/^.*?(\d+).*$/$1/s Beware of that if there is no digit in $file, then $sk won't change! This is bad. It would look like you need to change \d+ to \d*, but that would make it match "" at the beginning, and so $1 would be empty, thus erasing the whole string.

      But this problem can be solved. By rewriting the pattern to a more natural (?) pattern we'll soon see the solution. First, your pattern can be rewritten to   s/\D*(\d+).*/$1/s The anchors are removed, as they're unnecessary. They're unnecessary in your pattern too. Anyhow, now we can change \d+ to \d*, and $sk will be empty if no number was matched. So the result is   (my $sk = $file) =~ s/\D*(\d*).*/$1/s; But this still isn't fully analogous with   my ($sk) = $file =~ /(\d+)/; since $sk will be the empty string in the former, and the undefined value in the latter. So in extraction situations I often stay away from that trick, and simply use the latter.

      ihb
[reply]
[d/l]
[select]
Re: Regex not behaving as expected
by pizza_milkshake (Monk) on Feb 01, 2003 at 12:15 UTC
    context. your example has $sk in scalar context. you need list context, i.e. ($sk)

    perl -e'$_=q#: 13_2: 12/"{>: 8_4) (_4: 6/2"-2; 3;-2"\2: 5/7\_/\7: 12m m::#;s#:#\n#g;s#(\D)(\d+)#$1x$2#ge;print'

[reply]
[reply]

        Not necessarily. If there are no capture groups then the list context result behaves as if there is an implicit group around the entire regex.

        Seeking Green geeks in Minnesota

[reply]

Back to Seekers of Perl Wisdom