PHP Sql Server Output Parameter

Question

I\'m struggling with an output parameter in PHP and SQL Server 2008. I keep getting the error:

Notice: Undefined variable: UserID in C:\\inetpub\\www

Answer 1

I am not 100% sure, but I believe this is asking you to assign a type to the output.
Try changing this:

$params = array(
                array($_POST["FirstName"], SQLSRV_PARAM_IN),
                array($_POST["LastName"], SQLSRV_PARAM_IN),
                array($_POST["Email"], SQLSRV_PARAM_IN),
                array($_POST["Username"], SQLSRV_PARAM_IN),
                array(md5($_POST["Password"]), SQLSRV_PARAM_IN),
                array(date("Y-m-d H:i:s"), SQLSRV_PARAM_IN),
                array($_SERVER["REMOTE_ADDR"], SQLSRV_PARAM_IN),
                array("Member", SQLSRV_PARAM_IN),
                array("No", SQLSRV_PARAM_IN),
                array($UserID, SQLSRV_PARAM_OUT)
               );

to

$params = array(
                array($_POST["FirstName"], SQLSRV_PARAM_IN),
                array($_POST["LastName"], SQLSRV_PARAM_IN),
                array($_POST["Email"], SQLSRV_PARAM_IN),
                array($_POST["Username"], SQLSRV_PARAM_IN),
                array(md5($_POST["Password"]), SQLSRV_PARAM_IN),
                array(date("Y-m-d H:i:s"), SQLSRV_PARAM_IN),
                array($_SERVER["REMOTE_ADDR"], SQLSRV_PARAM_IN),
                array("Member", SQLSRV_PARAM_IN),
                array("No", SQLSRV_PARAM_IN),
                array($UserID, SQLSRV_PARAM_OUT, SQLSRV_PHPTYPE_INT)
               );

ref from: here