轉換AWK對Perl

Question

我無法搞清楚如何這個簡單的AWK命令轉換成Perl

awk -F: '{if($3 > 999 && $3 < 30000){ print $1}}' /etc/passwd 

的思考？

Answer 1

#!/usr/bin/perl   

use strict; 
use warnings; 

while (1) { 
    my ($user, $uid) = (getpwent())[0,2] 
     or last; 

    print("$user\n") 
     if $uid >= 1_000 && $uid < 30_000; 
} 

endpwent(); 

Answer 2

字面相當於是這樣

perl -aF: -ne'print "$F[0]\n" if $F[2] > 999 and $F[2] < 30_000' /etc/passwd