Something like this, perhaps:

#!/usr/bin/perl
use strict;
use warnings;

my %replace = (
    Not         => 'Just',
    Stupid      => 'Perl',
);

my $string = "[Not] Another [Stupid] Hacker";

$string =~ s/\[([^\]]+)\]/$replace{$1}/g;

print $string, $/;

__OUTPUT__
Just Another Perl Hacker
[download]

my %replacement = (
  A => 'B',
  C => 'D',
);

my $loop_line = '[A]A[C]C';

$loop_line =~ s/\[([^\]]+)\]/$replacement{$1}/g;

print "$loop_line\n";
[download]

Thank you all that replied. Especially so quickly. My problem is now solved - I really should master regex one day as it is often where I have trouble. Thanks again.......

#!/usr/bin/perl

use strict;
use warnings;

use Benchmark;

our $loop_line = "static replace [something] and [another] thing";

our %replace = (something => "SOMETHING",
                another   => "ANOTHER");

my ($a, $b, $c);

timethese -1 => {
    lazy         =>  '$a = $loop_line; $a =~ s/\[(.*?)\]/$replace{$1}/
+g',
    restrictive  =>  '$b = $loop_line; $b =~ s/\[([^]]*)\]/$replace{$1
+}/g', 
    copy         =>  '$c = $loop_line;',
};

__END__
Benchmark: running copy, lazy, restrictive for at least 1 CPU seconds.
+..
       copy:  2 wallclock secs @ 3994936.79/s (n=4234633)
       lazy:  1 wallclock secs @  121663.37/s (n= 122880)
restrictive:  1 wallclock secs @  117028.57/s (n= 122880)
[download]

Abigail

Well the first problem you are likely having is that with your regex, when you have something like static replace [something] and [another] thing it is grabbing the longest it can and therefore $replace ends up being something] and [another. A simple character class along the lines of [^]] (grabs everything but the closing bracket) instead of just . should do it

The second problem I see is that when the replacement is done, it leaves the brackets there, so if you do iterate over it again, it will proceed to attempt to replace the replacements, which will likely not have any corresponding value in the hash, and therefore be undefined, and leave the spot empty. Adding escaped brackets into the the first half of the substitution will take care of that problem

So, after all that here is my test case which should demonstrate one way to do this with a while loop and hashes as you first mentioned:

$loop_line = "static replace [something] and [another]\n";
%replace_table = ('something','this','another','that');
print $loop_line;

while ($loop_line =~ /\[([^]]*)\]/) {
    my $replace = $1;
    my $replacement = $replace_table{$1};
    $loop_line =~ s/\[$replace\]/$replacement/;
    };

print $loop_line;
[download]

update: bleh, took too long writing this, so I got beaten to it :( ahh well, if nothing else I got some practice on RegExs and Debugging :)

my @matches = $loop_line =~/\[(.*?)\]/; # Build a list of matches
for ( @matches ) {
    $loop_line =~ s/\Q$_\E/$replace{$_}/; # \Q makes sure no strange t
+hings happen
}
[download]

Rubbish.

#!/usr/bin/perl

use strict;
use warnings;

my $loop_line = "static replace [something] and [another] thing";

my @matches = $loop_line =~/\[(.*)\]/; # Build a list of matches
for ( @matches ) {
    $loop_line =~ s/\Q$_\E/PING/;
}

print $loop_line, "\n";

__END__
static replace [PING] thing
[download]

The problem with the given code is the use of .*, which your solution doesn't solve at all. In fact, your solution doesn't address any problem. It's just adding slowness. It's worse than not giving an answer.

Abigail

/Off Topic/
I don't see how this reply must have helped alongwor with his problem. Mine at least pointed him somewhat in the right direction. We all make errors, I even dare to say you make them too. But I thought that this website is all about learning, so you should make errors.

Thanx.
Update: I'm sorry for misreading, and thereby not suppling the answere. But I still do not agree with Abigail's reply.
Hmm, I guess my post was rubbish after all, forget what I have posted here.
Will do better next time.
---
Berik