my %replacement = (
    500     => '000',
    988     => '000',
    1210    => '1000',
    1300    => '1000',
    2134    => '2000,
    2500    => '2000',
);

my $regex = join '|', keys %replacement;

$str =~ s/\b($regex)\b/$replacment{$1}/;
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Untested, but it should work pretty much like this.

Funny, exact, to the point -> ++!

---

s$$([},&%#}/&/]+}%&{})*;#$&&s&&$^X.($'^"%]=\&(|?*{%
+.+=%;.#_}\&"^"-+%*).}%:##%}={~=~:.")&e&&s""`$''`"e

...now all we need is a script to generate that %replacement hash :o)

-- FloydATC

Time flies when you don't know what you're doing

How about this?

use strict;
use warnings;
use Data::Dumper;

my $post = "Hi, erm does anyone know how to do a string replacement th
+at caters for all the following scenario? 500 changes to 000 or 0 988
+ changes to 000 or 0 1210 changes to 1000 1300 changes to 1000 2134 c
+hanges to 2000 2500 changes to 2000 Thanks in advance.";

my %replacement = $post =~ /(\d+) changes to (\d+)/g;

print Dumper \%replacement;
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What you describe, sounds like rounding down to the next 1000s. Have a look at this example:

use strict;
use warnings;

my @numbers = ( 500, 988, 1210, 1300, 2134, 2500 );

# rounding numerically
for (@numbers) {
    my $n = int( $_/1000 )*1000;
    print "$_ => $n\n";
}

# rounding in regex
for (@numbers) {
    s{(\d+)}{int( $1/1000 )*1000}e;
    print "$_\n";
}
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Without punctuation and string delimiters, it's very difficult to know exactly what you're asking. The guidelines in "How do I post a question effectively?" will help you post a better question.

Regex substitution is performed with s///. You should probably read "perlrequick - Perl regular expressions quick start" and "perlretut - Perl regular expressions tutorial" to get a basic understanding of Perl's regular expressions before doing anything else.

It sounds like you probably want a lookup table and a regex using alternation. Here's a very simplistic example of the technique:

#!/usr/bin/env perl -l

use strict;
use warnings;

my %lookup = ('500' => '000', '0 988' => '000', '0 1210' => '1000');
my $re = '(' . join('|' => keys %lookup) . ')';

my $string = 'www "999" xxx "0 1210" yyy "500" zzz "0 988"';
print $string;

$string =~ s/$re/$lookup{$1}/g;
print $string;
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Output:

www "999" xxx "0 1210" yyy "500" zzz "0 988"
www "999" xxx "1000" yyy "000" zzz "000"
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Depending on the context of the strings to be replaced, the pattern may need to be a lot more complex than that shown. For instance, given '500 988', do you want to replace 500' and ignore ' 988' or ignore '50' and replace '0 988'. Reading the documentation I linked to will help you understand these issues and how to deal with them. And, of course, assuming '0 988' is a string to be replaced is just a guess on my part (as I alluded to in the opening paragraph).

How would you determine whether 500 changes to 000 or 0 ? When should it change to 0?

I assume you just want to "nullify" the last 3 digits of your numbers. You could do it with the following code:

my @strings=("500", "988", "1210", "1300", "2000");

foreach my $in (@strings)
{
    print "$in ";
    $in =~ s/(\d\d\d)$/000/;  # take the last 3 digits, replace them w
+ith 000
    print "-> $in\n";
}
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