Regex, PHP - finding words that need correction -

i have long string words. of words have special letters.

for example string "have rea$l problem with$ dolar inp$t" , have special letter "$".

i need find , return words special letters in quickest way possible.

what did function parse string space , using “for” going on words , searching special character in each word. when finds it—it saves in array. have been told using regexes can have better performance , don’t know how implement using them.

what best approach it?

i new regex understand can me task?

my code: (forbiden const) code works now, 1 forbidden char.

function findspecialchar($x){ $special = ""; $exploded = explode(" ", $x); foreach ($exploded $word){    if (strpos($word,$forbidden) !== false)      $special .= $word;  } return $special; } 

you use preg_match this:

// set special word here. $special_word = "café";  // set sentence here. $string = "i eat food @ café , read magazine.";  // run through 'preg_match''. preg_match("/(?:\w|^)(\q$special_word\e)(?:\w|$)/i", $string, $matches);  // dump results see working. echo '<pre>'; print_r($matches); echo '</pre>'; 

the output be:

array (     [0] =>  café     [1] => café ) 

then if wanted replace that, using preg_replace:

// set special word here. $special_word = "café";  // set special word here. $special_word_replacement = " restaurant ";  // set sentence here. $string = "i eat food @ café , read magazine.";  // run through 'preg_replace''. $new_string = preg_replace("/(?:\w|^)(\q$special_word\e)(?:\w|$)/i", $special_word_replacement, $string);  // echo results. echo $new_string; 

and output be:

i eat food @ restaurant , read magazine.

i sure regex refined avoid having add spaces before , after " restaurant " in example, basic concept believe looking for.