Yes, it's not perl ;-)

#!/usr/bin/perl
use strict;
use warnings;
my $uid = $ARGV[0];

open (my $fh, '<', '/etc/passwd') or die "Can't open passwd file: $!";

while (<$fh>){
    my $this_uid = (split m/:/)[2];
    print if $this_uid > $uid;
}
close $fh;
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You should try comp.unix.shell newsgroup for shell, including korn shell (or simply ksh), questions instead of this site related to things Perl.

In any case ...

#!/usr/bin/ksh
uid=${uid:-/etc/passwd}
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Why not just set $uid directly (uid='/etc/passwd')?

/etc/passwd/read
$ cut-d: -f1 < uid gt $$
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Just what are you trying to do there? Does your script run to completion successfully (not that it gives a result)?

export uid=1000; perl -ne '$1 > $ENV{"uid"} and print if /\w+:\w+:(\d+
+)/' /etc/passwd
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