CODEWITHSUNDEEP
regexperl
Dana
Dana

Reputation: 19

How to extract in perl just a part of a string?

I have a variable that contains a path like this:

$Path = "C:/Work/Job_V14/Myfile";

This variable can change each time I run the code. The only thing that never changes is that in the Path I will always have _V and 2 digit (like in this case _V14).

How can I obtain from the path only the part _V+2 digit?

Upvotes: 0

Views: 100

Answers (4)

Polar Bear
Polar Bear

Reputation: 6818

Please see the following piece of code

#!/usr/bin/perl

use strict;
use warnings;
use feature 'say';

while( <DATA> ) {
    say $1 if m!(_V\d{2})/!;
}

__DATA__
C:/Work/Job_V14/Myfile
C:/Work/Project_01/Job_V12/SomeFile
C:/Work/Project_03/Job_V02/SomeFile
C:/Work/Job_V34/SomeFile

Output:

_V14
_V12
_V02
_V34

Upvotes: 0

Patrick Taylor
Patrick Taylor

Reputation: 71

$Path = "C:/Work/Job_V14/Myfile";

# this code makes the assumption that your _Vxx contains the first _ in the path string
my $first; my $second; my $result ;

($first, $second) = split('_',$Path,2);   # split off the part before _
($result) = split('/',$second) ;   # takes only the first element 
$result = "_" . $result ; #adding back the _

print $result . "\n" ;  # prints _V14

If you have multiple occurrences of _ before your _Vxx, then this code needs an adjustment.

Upvotes: 1

pmqs
pmqs

Reputation: 3735

You can use a regular expression to match the substring, like so

$Path = "C:/Work/Job_V14/Myfile";

if ( $Path =~ m[ ( _V \d\d [/\\] ) ]x )
{
    my $match = $1 ;
    print "Matched '$match'\n";
}
else
{
    print "No match\n";
}

Above assumes that _V+2digit can never appear anywhere else in the string. If that is the case, the regular expression needs to be revisited.

Upvotes: 1

ikegami
ikegami

Reputation: 386706

If the path will always be a file in the directory whose name you want, I'd start with the following:

use Path::Class qw( file );

my path = "C:/Work/Job_V14/Myfile";
my $project = file($path)->dir->basename;   # Job_V14

Now, we us a match to extract the version.

my ($version) = $project =~ /_V(\d+)\z/;    # 14

Upvotes: 2

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