Complex(?) Name Matching Regex for vBulletin

Question

I'm creating some custom BBcode for a forum. I'm trying to get the regular expression right, but it has been eluding me for weeks. Any expert advice is welcome.

Sample input (a very basic example):

[quote=Bob]I like Candace. She is nice.[/quote]

Ashley Ryan Thomas

Essentially, I want to encase any names (from a specified list) in [user][/user] BBcode... except, of course, those being quoted, because doing that causes some terrible parsing errors.

The desired output:

[quote=Bob]I like [user]Candace[/user]. She is nice.[/quote]

[user]Ashley[/user] [user]Ryan[/user] [user]Thomas[/user]

My current code:

 $searchArray = array(
 '/(?i)([^=]|\b|\s|\/|\r|\n|\t|^)(Ashley|Bob|Candace|Ryan|Thomas)(\s|\r|\n|\t|,|\.(\b|\s|\.|$)|;|:|\'|"|-|!|\?|\)|\/|\[|$)/'
 );

 $replaceArray = array(
   "\\1[user]\\2[/user]\\3"
 );

 $text = preg_replace($searchArray, $replaceArray, $input);

What it currently produces:

 [quote=Bob]I like [user]Candace[/user]. She is nice.[/quote]

 [user]Ashley[/user] Ryan [user]Thomas[/user]

Notice that Ryan isn't encapsulated by [user] tags. Also note that much of the additional regex matching characters were added on an as-needed basis as they cropped up on the forums, so removing them will simply make it fail to match in other situations (i.e. a no-no). Unless, of course, you spot a glaring error in the regex itself, in which case please do point it out.

Really, though, any assistance would be greatly appreciated! Thank you.

Answer 1

Since you don't really need to match the spaces on either side (just make sure they're there, right?) try replacing your search expression with this:

$searchArray = array(
  '/\b(Ashley|Bob|Candace|Ryan|Thomas)\b/i'
);

$replaceArray = array(
 '[user]$1[/user]'
);

$text = preg_replace($searchArray, $replaceArray, $input);

Answer 2

It's quite simply that you are matching delimiters (\s|\r|...) at both ends of the searched names. The poor Ashley and Ryan share a single space character in your test string. But the regex can only match it once - as left or right border.

The solution here is to use assertions. Enclose the left list in (?<= ) and the right in (?= ) so they become:

 (?<=[^=]|\b|\s|\/|^)
 (?=\s|,|\.(\b|\s|\.|$)|;|:|\'|"|-|!|\?|\)|\/|\[|$)

Btw, \s already contains \r|\n|\t so you can probably remove that.