Help with regex code

areid

my code is not working for some reason. Here is the code. If you see anything wrong, please let me know.

Thanks

$oURL = '<span class="text2"><a href="javascript:##openURL(\'{URL}\')">{DESCRIPTION}</a></span>';

$openURL = str_replace('{URL}', '\\1', $oURL);
$openURL = str_replace('{DESCRIPTION}', '\\2', $openURL);


$patterns = array();
$replacements = array();

$patterns[] = "#\[openURL=\"(.*?)\"](.*?)\[/openURL\]#is";
$replacements[] = $openURL;

$text = preg_replace($patterns, $replacements, $text);

##openURL is really openURL without the '##'

this is an example of whats in the database:

[openURL=../includes/video_tape.mtm]details[/openURL]

bodzan

That the error is in this line:

$oURL = '<span class="text2"><a href="java script:##openURL('{URL}')">{DESCRIPTION}</a></span>';

Since there are apostrophes inside this string beside {URL}... Try escaping them, but I'm not 100% positive that it is the main problem of your code...

areid

Still the same thing. I think the error is comming from here:

$patterns[] = "#\[openURL=\"(.*?)\"](.*?)\[/openURL\]#is";

But I dont know what it is

areid

Ok got it working. Thanks for you sugjestion!

$oURL = '<span class="text2"><a href="javascript:##openURL(##\'{URL}##\')">{DESCRIPTION}</a></span>';

$openURL = str_replace('{URL}', '\\1.\\2', $oURL);
$openURL = str_replace('{DESCRIPTION}', '\\3', $openURL);

$patterns = array();
$replacements = array();

$patterns[] = "#\[openURL=([^ \"\n\r\t<]*?)\.([^ \"\n\r\t<]*?)\]([^ \"\n\r\t<]*?)\[/openURL\]#si";
$replacements[] = $openURL;

$text = preg_replace($patterns, $replacements, $text);

Ignore the double number signs '##' The board messes it up