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

...and probably is. I'm trying to filter out any non-(alphanumeric or whitespace) characters from a string. 
$string !~ s/\w|\S//g;

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(or anything like it that I tried) doesn't work and the few tangles with look-aheads have proven unsuccessful.

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RE: It seems simple...
by Anonymous Monk on Feb 25, 2000 at 13:23 UTC
    by this you mean: "ab&^#c( 1 2"=>"abc12"? if so try $string=~s/\W//g; not sure what you meant, the $var!~s///g syntax seems weird to me.
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Re: It seems simple...
by Anonymous Monk on Feb 24, 2000 at 09:21 UTC
    The !~ operator negates the return value of the =~ operator, not of the regular expression. In an s/// operation, the result of $string will be the same regardless of whether !~ or =~ is used, as all that is being negated is the number of matches returned. What you're looking for is $string =~ s/\W|\s//g;
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      Unfortionatly, that doesn't work. 
      s/\W|\s//g;

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      successfully does what it says it will - both: 
      s/\W//g;

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      and 
      s/\s//g;

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      thereby eliminating the whitespace that I wanted to preserve (both regular expressions do that, actually)
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        Sorry, I misread your question the first time (me having posted the first answer).. I think what you want is $string =~ s/[^\w\s]//g;
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