Badwords filter, finding only "whole word"

Use regex with preg_replace, if you want to censor the words out.  Or use preg_match to just get a true/false if the word is in the haystack.

http://php.net/manual/en/function.preg-replace.php
http://php.net/manual/en/function.preg-match.php

Thanks
But regex is not my forte
could you show me the way with preg_replace?
regards

Read this:
http://bytes.com/topic/php/answers/883266-preg_replace-whole-word-only

Like this.

<?php
$string = 'Experts Exchange.';
$pattern = '/cha/';
$replacement = '###';
echo preg_replace($pattern, $replacement, $string);

//Echos  Experts Ex###nge
?>

Select all Open in new window

You can also do it as an array

<?php
$string = 'Experts Exchange has a lot of answers for questions.';
$patterns = array();
$patterns[0] = '/answers/';
$patterns[1] = '/for/';
$patterns[2] = '/questions/';
$replacements = array();
$replacements[0] = 'problems';
$replacements[1] = 'to';
$replacements[2] = 'solve';
echo preg_replace($patterns, $replacements, $string);
//Echos  Experts Exchange has a lot of problems to solve.
?>

Select all Open in new window

Thanks,
but it does not resolve my quest, the first solution is getting the same result as what I do know.
I am looking for it not to pick up "cha" in exchange but to pick it up in a str like: this is cha a str"
as a whole word only

And the other solution I am not up to adapt it
as I did not find a way to make my array working as the pattern
it looks like:
<?php
$badwords =
array (
'zzzzz',
etc.....

Thanks works a treat!
I just needed, as I got a var error,
to declare before the concat
wordlist as empty
$wordlist='';