Find where and which word matched sql LIKE statement

Question

I am using a very long like query to detect offensive words within my users usernames.

I have made a moderator page where the moderator can click a button next to the detected username and it will replace the offending word with *'s. The problem is how can I easily detect which offensive word and where to replace?

I guess I would need to use a regex of some description that functions exactly the same as the LIKE statement. Help in creating such a regex would be greatly appropriated or... if anyone has a better solution.

TIA

Answer 1

You can try below functions in php

similar_text()
levenshtein()
soundex()

You can just create a simple script to match words and then replace those words as per your need.

Check for documentation http://in2.php.net/manual/en/function.similar-text.php

Answer 2

Something like:

$bad_list = array('foo','barr','ax'); // list of bad words you want to *

$from = array();
$to = array();

foreach($bad_list as $s){
    $from[] = '/'.preg_quote($s,'/').'/'; // PHP expects regex in delimiter.
    $to[] = str_repeat('*',strlen($s));
}
$str = "afoob";
$str = preg_replace($from,$to,$str); // $str is now a***b

You might also use str_replace, there is really no need of regex here.