Perl associative arrays and variable assignment

Question

I'm having trouble understanding how this is supposed to work. I have defined my two hashes outside of the while loop. And the goal is to be able to detect the shell type, and assign the appropriate hash to a new $shell variable. Here is the code I'm current trying..

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

use POSIX;
use DateTime;
use Term::ANSIColor qw(:constants);

local $Term::ANSIColor::AUTORESET = 1;
my $dt = DateTime->now;   # Stores current date and time as datetime object

my %multicity = (
        url        => "example.com",
        ftpuser    => "user",
        ftppass    => "pass",
        remote_dir => "/httpdocs/",
        dbname     => "database"
);

my %singlecity = (
        url        => "example2.com",
        ftpuser    => "user",
        ftppass    => "pass",
        remote_dir => "/httpdocs/",
        dbname     => "database"
);

open (MYFILE, 'sites.txt');

LINE: while (<MYFILE>) {
    next LINE if /^#/;
    my ($shelltype, $siteurl, $ftpuser, $ftppass, $dbname) = split /\|/;

    # 1|example.com|user|pass|databasename - This should be a singlecity shell type.
    if ($shelltype == 1) { my $shell = \%multicity; } else { my $shell = \%singlecity; };

    print "Shelltype: $shelltype\n";
    print "Should be $shell{url}\n";

}
close (MYFILE);

I've tried many different things with no avail so I'm finally resorting to the pros here at stackoverflow to get some direction!

Any help is very much appreciated. Thank you.

Answer 1

Your my $shell is lexical inside of the if block. Move it outside the if or it is not available there. Adding some indentation helps to spot that.

my $shell;
if ($shelltype == 1) { 
  $shell = \%multicity; 
} else { 
  $shell = \%singlecity; 
};

After that, you will get a warning saying %shell is undefined. That's because you are using $shell{url} but you defined $shell as a hash ref, so you need $shell->{url} or $$shell{url} in your print.

Answer 2

Two issues:

1. Your my $shell variable is lexically scoped to the if statement. It won't be visible outside.

2. The $shell variable is a hashref. So you have to access elements using $shell->{url}. Note the arrow.

So the following should work:

my $shell;
if ($shelltype == 1) { $shell = \%multicity; } else { $shell = \%singlecity; };

print "Shelltype: $shelltype\n";
print "Should be $shell->{url}\n";

Answer 3

This line:

print "Should be $shell{url}\n";

should be:

print "Should be $shell->{url}\n";

The reason is that you are assigning a reference to one of your two hashes to $shell, so to access a value in it you need to de-reference (using the -> operator).

In addition, as pointed out by nwellholf, the $shell variable is local to the if statement so you would get a compilation error there.