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

I am wondering if it is all possible to consolidate the following example into a single regular expression instead of using two

if ( /^[^0-9]+$/ ) {
   my @words= /(\b.+?\b)/g;
   print "@words\n";
}

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All this does is match words as long as there is not a single number located anywhere in the line.

Replies are listed 'Best First'.
Re: Reg Ex help
by Chmrr (Vicar) on Sep 12, 2001 at 21:35 UTC
    How about the following:
    if (my @words = /(\b.*?\b)/g and not grep {/\d/} @words) {
  print @words;
  # Do stuff.
}

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    However, having two regex isn't always a bad thing. Personally, I think your code reads easier and is more intuitive; my only change would be from if (/^[^0-9]+$/) to unless (/\d/)

     
    perl -e 'print "I love $^X$\"$]!$/"#$&V"+@( NO CARRIER'

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Re: Reg Ex help
by chromatic (Archbishop) on Sep 12, 2001 at 21:33 UTC
    Depending on what you consider a word, this may be a better solution: 
    unless (/\d/) {
    @words = split(' ', $_);
}

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Re: Reg Ex help
by dragonchild (Archbishop) on Sep 12, 2001 at 20:30 UTC 
    unless (my @words = /(\b\D+?\b)/g) {
    print "Houston, we have a problem.\n";
}

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    ------
    We are the carpenters and bricklayers of the Information Age.

    Don't go borrowing trouble. For programmers, this means Worry only about what you need to implement.

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      That doesn't work. He doesn't want to match at all if theres any digits anywhere in the string (or so he says :). Try "abc 123" just for example. Sometimes its just not worth trying to do something in one regex, but you could simplify to this:
      unless (tr/0-9//) {
 my @words = /\w+/g;
 print "@words\n";
}

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        No, it's not identical. But, I suspect, it's closer to the spirit of what he's doing...

        ------
        We are the carpenters and bricklayers of the Information Age.

        Don't go borrowing trouble. For programmers, this means Worry only about what you need to implement.

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Re: Reg Ex help
by Anarion (Hermit) on Sep 13, 2001 at 05:31 UTC

    I think you want something like:

    @words=grep{!/\d/}split;

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    $anarion=\$anarion;

    s==q^QBY_^=,$_^=$[x7,print

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Re: Reg Ex help
by jehuni (Pilgrim) on Sep 13, 2001 at 20:52 UTC
    !/\d/ and @words = /(\b.+?\b)/g and print "@words\n";

    -jehuni

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Re: Reg Ex help
by hopes (Friar) on Sep 14, 2001 at 09:41 UTC
    How about that:

    @words=grep{$_}/(?:.*\d.*)?(\b\w+?\b)?/g;

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    Note that the anonymous monk want to match anything if there is a digit in the string, and he want to use only a reg exp.
    I had problems, because I was matching '' (empty).
    The grep solves that.

    I hope this help.

    P.S. I think it would be better with two reg exp's, but we know, TIMTOWTDI

    Regards, Hopes
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Re: Reg Ex help
by pike (Monk) on Sep 14, 2001 at 15:02 UTC
    How about this:

    my (@words) = /^\s*(\D+\b)+\s*$/;

    Just one regex, and should not match if the line contains digits.

    pike

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