http://www.perlmonks.org?node_id=1033760

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

I got a string $ou "# PATH=$PATH:$NAS_DB/sbin:$NAS_DB/bin:$NAS_DB/sbin" and want to delete the heading # and the white space around.

 my $ou ="# PATH=\$PATH:\$NAS_DB/sbin:\$NAS_DB/bin:\$NAS_DB/sbin";
print __FILE__." line".__LINE__." trace: ".$ou."\n";
if($ou =~ /^\s*#\s*(PATH=.*\$NAS_DB\/bin.*)$/)
{
my $str = $1;
$ou=~s/^\s*#\s*PATH=.*\$NAS_DB\/bin.*$/$str/;
print __FILE__." line".__LINE__." trace: ".$ou."\n";
}

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When I remove $str, replace it with $1, the output is empty. Not what I want. So why? Every thanks.

 my $ou ="# PATH=\$PATH:\$NAS_DB/sbin:\$NAS_DB/bin:\$NAS_DB/sbin";
print __FILE__." line".__LINE__." trace: ".$ou."\n";
if($ou =~ /^\s*#\s*(PATH=.*\$NAS_DB\/bin.*)$/)
{
$ou=~s/^\s*#\s*PATH=.*\$NAS_DB\/bin.*$/$1/;
print __FILE__." line".__LINE__." trace: ".$ou."\n";
}

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OK, I got it. Seems $1 was reset after every matching.

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Re: regex substitute(resolved)
by Athanasius (Archbishop) on May 16, 2013 at 04:13 UTC
    Seems $1 was reset after every matching.

    Correct. The pragma use warnings would tell you:

    Use of uninitialized value $1 in substitution iterator at...

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    which indicates that $1 is undef, because nothing has been captured in the current substitution. Easily fixed by adding parentheses to capture the text you want to keep:

    $ou =~ s/^\s*#\s*(PATH=.*\$NAS_DB\/bin.*)$/$1/;

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    Hope that helps,

    Athanasius <°(((>< contra mundum Iustus alius egestas vitae, eros Piratica,

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[select]
      Yes, fine. Thank you!
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Re: regex substitute(resolved)
by hdb (Monsignor) on May 16, 2013 at 06:29 UTC

    Unless you really need to confirm the structure of the path itself, the easiest way to remove the leading # and the surrounding whitespace would be: 

    $ou =~ /^\s*#\s*//;

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