Question

$str = preg_replace('/[\\r\\n!.,\'“”;’?-\s+]/', ' ', $str);
$str=preg_replace('/\s+\S{1,2}(?!\S)|(?<!\S)\S{1,2}\s+/', '', $str);

If I leave them like that is there performance issue also or is it better.?

Answer 1

preg_replace accepts arrays as arguments. The following should be equivalent:

$str = preg_replace(array('/[\\r\\n!.,\'“”;’?-\s+]/', '/\s+\S{1,2}(?!\S)|(?<!\S)\S{1,2}\s+/'), array(' ', ''), $str);