If you numerically sort your filenames in descending order, you will have the biggest in the first element (index 0) and the next biggest in the next element.

Modifying the code I provided earlier: 

#!/usr/bin/perl -l

use strict;
use warnings;
use autodie;

my $dir = '.';

opendir(my $dh, $dir);
my @emptyfiles = sort { $b <=> $a } grep { -f && /^\d+$/ && -z } readd
+ir $dh;
closedir $dh;

print for @emptyfiles;
print "Biggest: $emptyfiles[0]";
print "Next biggest: $emptyfiles[1]";

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Gives this output: 

123
1
Biggest: 123
Next biggest: 1

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So, they're effectively in separate variables already (i.e. $emptyfiles[0] and $emptyfiles[1]). There's probably no need to use other variables but, if you really need to, just do something like this (untested): 

my ($biggest, $next_biggest) = @emptyfiles[0, 1];

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Regarding your problem with "It keeps saying that $previous is -1", there's at least two problems. Firstly, the code you posted 

$previous = ($emptyfile)-1);

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won't compile. You'll get something like: "syntax error at ... line ..., near "1)""

So, the code you posted isn't the code you're running; however, given you don't initialise $emptyfile, subtracting 1 from it will evaluate to -1 and should also give you a warning (assuming you still have use warnings; at the top of your code). Here's a minimal example: 

$ perl -wE 'my $x; say($x - 1)'
Use of uninitialized value $x in subtraction (-) at -e line 1.
-1

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— Ken

In reply to Re^2: create array of empty files and then match filenames by kcott
in thread create array of empty files and then match filenames by angela2

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