Using a variable as a directory name in perl

Question

I am using a user input variable as my initial directory name, but when trying to expand the directory and create sub-folders, utilizing that variable causes issue in the path.

#!/usr/bin/perl
use strict;
use warnings;

print "What would you like to name your directory?\n"; 
chomp( my $directory = <STDIN> );

mkdir $directory, 0755;
mkdir $directory/data, 0755; ##<----<<Error begins here##
mkdir $directory/data/image, 0755;
mkdir $directory/data/cache, 0755;

Kudos for use strict; and use warnings;. Tiz nice to see a new user having already learned that most important lesson. — Miller
– Miller, Commented Sep 26, 2014 at 20:44

choroba · Accepted Answer · 2014-09-26 13:41:40Z

4

Unquoted / in Perl means division or regular expression match. Quote it if it's part of a string:

mkdir "$directory/data", 0755;

answered Sep 26, 2014 at 13:41

choroba

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Miller · Accepted Answer · 2014-09-26 20:43:11Z

3

Or try

use File::Path qw/ make_path /;

make_path( "${directory}/data/image", "${directory}/data/cache" );

it will create the intermediate directories for you.

You may also want to look at the Path::Tiny or Path::Class modules, which have a nicer OO-interface for file operations.

edited Sep 26, 2014 at 20:43

Miller

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answered Sep 26, 2014 at 14:00

Rob

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