Replace all spaces which are enclosed within braces

Question

What I want to do is find all spaces that are enclosed in braces, and then replace them with another character.

Something like:

{The quick brown} fox jumps {over the lazy} dog

To change into:

{The*quick*brown} fox jumps {over*the*lazy} dog

I already searched online, but only this is what I got so far, and it seems so close to what I really want.

preg_replace('/(?<={)[^}]+(?=})/','*',$string);

My problem with the above code is that it replaces everything:

{*} fox jumps {*} dog

I was looking into regexp tutorials to figure out how i should modify the above code to only replace spaces but to no avail. Any input will be highly appreciated.

Thanks.

Answer 1

Assuming that all braces are correctly nested, and that there are no nested braces, you can do this using a lookahead assertion:

$result = preg_replace('/ (?=[^{}]*\})/', '*', $subject);

This matches and replaces a space only if the next brace is a closing brace:

(?=     # Assert that the following regex can be matched here:
 [^{}]* #  - Any number of characters except braces
 \}     #  - A closing brace
)       # End of lookahead

Answer 2

I am reacting to your comment that you dont want to use regex, just string manipulation. That's OK but why have you then written that you are looking for a regex?

Solution wihout regex:

<?php

$str = "{The quick brown} fox jumps {over the lazy} dog";

for($i = 0, $b = false, $len = strlen($str); $i < $len; $i++)
{ 
    switch($str[$i])
    {
        case '{': $b = true; continue;
        case '}': $b = false; continue;
        default:
        if($b && $str[$i] == ' ')
            $str[$i] = '*';
    }
}

print $str;

?>

Answer 3

How about this:

$a = '{The quick brown} fox jumps {over the lazy} dog';
$b = preg_replace_callback('/\{[^}]+\}/sim', function($m) {
    return str_replace(' ', '*', $m[0]);
}, $a);
var_dump($b); // output: string(47) "{The*quick*brown} fox jumps {over*the*lazy} dog"