Re: how to print only 47 and another number will rejected and when user give 47 that number only print ?

[Kenosis]

Here's another option:

use strict;
use warnings;

my $num;

do {
    chomp( $num = <> );
} until $num eq '47';

print $num;

This keep's doing the code block (getting input from STDIN) until $num is 47. $num's printed only if it's 47. Why a eq instead of a ==? If 'asdf' is entered when using a ==, you'll get an Argument "asdf\n" isn't numeric in numeric eq (==) at ... messsage, but not if eq is used.

Did you want stealth? The following will not echo anything to the screen until 47 is entered, and then 47 will be printed:

use strict;
use warnings;
use Term::ReadKey;

my $num;
ReadMode 'noecho';

do {
    chomp( $num = ReadLine(0) );
} until $num eq '47';

ReadMode 'restore';
print $num;

Hope this helps!