in reply to Validating Numbers in CGI Script?

the following regex looks for a non-numeric character (if it fails then they are all digits - and therefore acceptable) in the $invalue string: 
my $invalue = param( 'INVALUE' );
if ( defined $invalue ) {
    if ( $invalue =~ /\D/ ) {
        $result = "$invalue is *not* a Number@"
    } else {
        $result = "$invalue is a number.";
    }
} else {
    $result = "Enter a value to test."
}

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The same could be achieved with $invalue =~ /^\d+$/ but as we all know... 
$less->{'characters'}->{'typed'} == `"time in the pub" | more`; # :-D

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hope this helps

Update: _doh_!! just re-read your post and realised I would have to quickly whip up a regex to allow "-"s and a possible "." so here it is... 

/^(?:-)?\d+(?:\.\d+)?$/

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oh, and for people who like to leave off a leading 0 on decimals > -1 but < 1... 
/^(?:-)?(?:\d+(?:\.\d+)?|\.\d+)$/

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I would be interested in an optimisation of this if anyone can do better.

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Re: Re: Validating Numbers in CGI Script?
by japhy (Canon) on May 10, 2001 at 18:58 UTC
    I know it's in the FAQ, but I will construct here on my own, for the good of all. 
    /^(?:-)?(?:\d+(?:\.\d+)?|\.\d+)$/

# becomes

/^-?(?:\d+(?:\.\d+)?|\.\d+)$/

# UPDATE!

/^-?(?=\d|\.\d)\d*\.?\d+$/

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    The look-ahead is used to remove the redundancy in your regex. It basically ensures there's either a digit, or a dot and THEN a digit. Then, it matches digits and/or a dot and digits, which we know will be valid.

    japhy -- Perl and Regex Hacker
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[d/l]
      Of course, if you're going to use a regex(p), you might as well make it easy on yourself:

      use Regexp::Common;

      if ($thing =~ /^$RE{num}{dec}$/) {
          ...
      }

      (this particular example, of course, requires DWIM.pm... :-)

      dha

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        I'm in favor of the user knowing what the regex does, before using some pre-packaged regex. No offense to Damian, of course, but I'd rather know what my regex does exactly. That said, some interested party could read the module's source.

        japhy -- Perl and Regex Hacker
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      I can only refer you to the article that MeowChow mentioned. Taking a look at Tom Christiansen's list of benefits that someone who rolls their own RE usually misses, you get zero out of 3 of them.

      RE's are not always the right answer...

      UPDATE
      Changed the link because the page moved. (Thanks footpad.)

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      cheers for that - don't know what I was thinking with the (?:-)? bit - maybe the smiley just cheered us all up!

      correct me if i'm wrong but won't your regex pick up a number which finishes a sentence _and_ the full stop (period to americans)? this would lead to perl treating it as a string in some cases (eg. ++ or --).

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        Hmm, actually, I think I meant to make the regex be /^-?(?=\d|\.\d)\d*\.?\d+$/. Yes, that looks far more sane. The other could have matched "1234.", which isn't proper. This one can match "1234", ".1234", and "12.34".

        japhy -- Perl and Regex Hacker
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        Well, let's take this guy apart. We know how the leading -? will be treated, so let's leave it out. Then we get two paths that this regex can follow:

        (?=\d)\d*\.?\d+$

        and

        (?=\.\d)\d*\.?\d+$

        If the regex takes the first path, we're guaranteed to get at least one digit. This digit must be present in the \d* or \d+ if it goes to the \d*, then we're guaranteed to match the equivalent of

        \d+\.?\d+

        which will match anything that's comprised of just numbers, and any two numbers with a single period between them.

        if it's in the \d+, then we know that there are no periods in the string, since we must have passed through the optional \d* and \.?

        so this first path will always match a number number with no fractional portion, or a fractional number with a mandatory leading whole number portion

        Now in the other path: (?=\.\d)\d*\.?\d+$

        If this regex matches, we're guaranteed to have a period then a number. Since this is a lookahead, that means we've seen past the following \d*, since we can't possibly have picked up a number prior to the period.

        So the actual regex in this path is \.\d+, which clearly just matches a number with no preceding 0.

        So in any case, I don't think it will match anything other than numbers, as japhy suggests.

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