note 61936 deleted from function.preg-replace by colder

Note Submitter: elaineseery at hotmail dot com ---- if you're new to this function, and getting an error like 'delimiter must not alphanumeric backslash ... note that whatever is in $pattern (and only $pattern, not $string, or $replacement) must be enclosed by '/ /' (note the forward slashes) e.g. $pattern = '/and/'; $replacement = 'sandy'; $string = 'me and mine'; generates 'me sandy mine' seems to be obvious to everyone else, but took me a while to figure out!!