http://www.perlmonks.org?node_id=1044655


in reply to Quick Regex trouble

Close; you need a negative look-behind assertion: 
#!/usr/bin/perl
use strict;
use warnings;

my $var = "use neighbor 2001:504:0:4::6181:1";
($var =~/(?<!use )neighbor[[:alpha:]-]* ([[:alnum:]\.-:]+)/) and do{
    print $1;
};

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#11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.

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Re^2: Quick Regex trouble
by 2teez (Vicar) on Jul 16, 2013 at 21:16 UTC

    NOTEI think it's actually look-behind assertion he needs. 

    ($var =~/(?<=use )neighbor[[:alpha:]-]* ([[:alnum:]\.-:]+)/) and do{
    print $1;
};

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    output 2001:504:0:4::6181:1 If that is the output the OP wanted.

    Because with "negative look-behind assertion", the match is not successful (failed). In fact, there is no output.
    Checked with use re 'debug';
    .. produces ..
    Read more... (3 kB)


    UPDATE:
    NOTE: I think I got this wrong, and kennethk was right! Since, the OP gave a condition on which the regex should match as "when the string doesn't have USE" and the sample he gave has it.

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      I think it's actually look-behind assertion he needs.

      I don't think so. The OP said that it should match "neighbor" but not "use neighbor", so that a zero-width negative look-behind assertion seems to be what is needed to exclude "use".

      Just an additional note: the do {} block is unnecessary in the OP code. The solution could be:

      ($var =~/(?<!use )neighbor[[:alpha:]-]* ([[:alnum:]\.-:]+)/) and print $1;

      or even:

      print $1 if $var =~/(?<!use )neighbor[[:alpha:]-]* ([\d:]+)/;
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