PHP preg_replace - finding the replacement from an array using the match as the key

Question

I have a string which can contain multiple matches (any word surrounded by percentage marks) and an array of replacements - they key of each replacement being the match of the regex. Some code will probably explain that better...

$str = "PHP %foo% my %bar% in!";
$rep = array(
  'foo' => 'does',
  'bar' => 'head'
);

The desired result being:

$str = "PHP does my head in!"

I have tried the following, none of which work:

$res = preg_replace('/\%([a-z_]+)\%/', $rep[$1], $str);
$res = preg_replace('/\%([a-z_]+)\%/', $rep['$1'], $str);
$res = preg_replace('/\%([a-z_]+)\%/', $rep[\1], $str);
$res = preg_replace('/\%([a-z_]+)\%/', $rep['\1'], $str);

Thus I turn to Stack Overflow for help. Any takers?

Answer 1

echo preg_replace('/%([a-z_]+)%/e', '$rep["$1"]', $str);

gives:

PHP does my head in!

See the docs for the modifier 'e'.

Answer 2

It seems that the modifier "e" is deprecated. There are security issues. Alternatively, you can use the preg_replace_callback.

$res = preg_replace_callback('/\%([a-z_]+)\%/', 
                             function($match) use ($rep) { return  $rep[$match[1]]; },
                             $str );