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


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