SELECT id
FROM lockers AS l
WHERE NOT EXISTS = (SELECT 1
                    FROM units AS u
                    WHERE u.locker = l.id)
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And here's a full demonstration:

($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
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Awesome! Thanks so much everyone who replied. This leads me in the right direction.

You could put your entire locker list into an array, put the leased lockers that you harvested from the DB into a hash, then generate a new array for locker numbers that have not been leased.

use warnings;
use strict;

my @lockers = (1..25);

my %leased = map {$_ => 1} (1..12, 14..25);

my @avail = grep {! defined $leased{$_}} @lockers;

print "$_\n" for @avail;

__END__
13
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You are looking for the disjunction between the set of all lockers and the set of rented lockers. So, in pseudo code is one way:

• I have an array of lockers, 1 .. 24, initialized to zeroes
• I have a database containing information
• Loop: Read a Record from the DB
• If the locker-id is present, set the value in my locker-array to 1.
• Go to Loop: until you exhaust the DB.
• scan the locker-array, if the value of the i-th element is 0, print the index (i).
• Turn the result into your Teacher, you have completed the assignment in a FORTRANly fashion.

L*

my %lockers = map { $_, 1 } 1..25;
delete @lockers{ @leased };
print "available @{[ keys %lockers ]}\n";
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#!/usr/bin/perl
my $id = 0 x 25;
my @leased = ( 17..25, 1..15 );# or however you set them, order not im
+portant
substr($id, $_ - 1, 1) = 1 for @leased;
$id =~ /0/ and print "available $+[0]\n";
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