Re^2: Use of uninitialized value $1

[Pazitiff]

Thanks! But what if i want capture the left word ("path_to_my_") of "lib"? In general, I need a regular expression to get what is left of the "lib". How this can be done using $1, $2, $3 ? Thanks. Sorry, but i have no experience in Perl and RX.

[AnomalousMonk]

The Devil is in the Details! Exactly what are you matching against, and exactly what do you want to match?

c:\@Work\Perl\monks>perl -wMstrict -le
"my $var = '%&!~ path_to_a_lib_of_my_lib';
 ;;
 $var =~ m{ (.*) lib }xms;
 print qq{A: '$1'};
 ;;
 $var =~ m{ (.*?) lib }xms;
 print qq{B: '$1'};
 ;;
 $var =~ m{ (\w*?) lib }xms;
 print qq{C: '$1'};
"
A: '%&!~ path_to_a_lib_of_my_'
B: '%&!~ path_to_a_'
C: 'path_to_a_'

Update: Please see perlre, perlretut, and perlrequick.

Further Update: If the string consists entirely in something like 'path_to_a_lib_of_my_lib', see also the start- and end-of-string "anchors"  \A \z \Z (particularly  \A \z in this case) for further narrowing what you want to capture.

[FreeBeerReekingMonk]

my $v = '/path/to/module.pm';
$v =~ m{^(.*)/(.*)$};
say "($1) ($2)";

prints out:

(/path/to) (module.pm)

Thus, if you want your module name hardcoded:

my $modulepath;
if($v =~ m{^(.*)module.pm$}){
 $modulepath = $1;
}

additional note: usually you write m// with slashes, but I write it with m{} brackets. This is also valid and allows you to use / inside the regexp, without having to escape it with a backslash.