mysqli_bind_result

Dear List - Here is some code:

            $sql13 = "SELECT * FROM Intake3 WHERE MedRec = ?";
            $stmt = mysqli_stmt_init($cxn);
            mysqli_stmt_prepare( $stmt, $sql13 );
            $_SESSION['stmt'] = $stmt;
            $args = array();

            $args[0] = &$_POST['MR'];
            $types = 'i';
            if( !$stmt )
                throw new Exception( 'Error preparing statement' );

            // Put the statement and types variables at the front of the params to pass to mysqli_stmt_bind_param()
            array_unshift($args, $stmt, $types ); // Note that I've moved this call. Apparently it doesn't pass back the result.
        //    I guess sometimes I just forget these things.

            // mysqli_stmt_bind_param()
            if( !call_user_func_array( 'mysqli_stmt_bind_param', $args ) )
            throw new Exception( 'Failed calling mysqli_stmt_bind_param' );

            if( !mysqli_stmt_execute( $stmt ) )
                throw new Exception( 'Error while executing statement' );

           mysqli_stmt_bind_result( $stmt,  $Site, $MedRec, $Fname, $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age);
                throw new Exception( 'Failed calling mysqli_stmt_bind_result' );

Fatal error: Uncaught exception 'Exception' with message 'Failed calling mysqli_stmt_bind_result' What is my error?? FYI - MR comes from here -- <center><strong>Do you Wish to Enter Visit Data?</center> <center>If Yes:</center>

		    	<form name="form2" method="post">

<center>Choose the Medical Record that You Wish to Use</strong></center> <form method="post"> <center><input type="text" name="MR"></input></center> <center><input type="submit" value="Enter Medical Record"/></center> <input type="hidden" name="next_step" value="step10" /> </form> Advice and help, please. Ethan