preg_replace表现不如预期？

Sup peeps. I got an issue here. I receive this data and just want to strip the <SOAP-ENV elements with their respective closing elements.

This is the header and body start part.

<?xml version="1.0" encoding="UTF-8"?>
<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:wsse="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd" xmlns:wsu="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd">
  <SOAP-ENV:Header></SOAP-ENV:Header>
  <SOAP-ENV:Body>
    <VisionDataExchange>

Now I run my regular expession on $xml the variable containing the entire xml data:

$xml = preg_replace("/<\\/?SOAP(.|\\s)*?>/",'',$xml);

Now my result is this. It actually stripped the openening tags but none of the closing tags? What am I missing here?

<?xml version="1.0" encoding="UTF-8"?>

  </SOAP-ENV:Header>

    <VisionDataExchange>

I suggest just matching everything inside a tag, not any character or whitespace. Have a look at this regex:

$re = "/<\\/?SOAP[^<>]+?>/"; 
$str = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>
<SOAP-ENV:Envelope xmlns:SOAP-ENV=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:wsse=\"http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd\" xmlns:wsu=\"http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd\">
  <SOAP-ENV:Header></SOAP-ENV:Header>
  <SOAP-ENV:Body>
    <VisionDataExchange>"; 
$subst = ""; 

$result = preg_replace($re, $subst, $str);

Ok, so after nearly breaking my skull on my desk, i found what the issue was. The regex is indeed working perfectly!there was a hidden \ in the string that caused the regex to fail.