I have a Function which returns the "# of New Friend Requests" that a User has, and am not sure what to return in case that there is a problem with my query in the Function.

Here is my code...
PHP Code:
    function getNewFriendRequestCount($dbc$requestee){
        
// Build query.
        
$q1 "SELECT COUNT(requestor)
                        FROM friend
                        WHERE requestee=?
                        AND requestor_approved=1 AND requestee_approved=0"
;

        
// Prepare statement.
        
$stmt1 mysqli_prepare($dbc$q1);

        
// Bind variable to query.
        
mysqli_stmt_bind_param($stmt1'i'$requestee);

        
// Execute query.
        
mysqli_stmt_execute($stmt1);

        
// Store results.
        
mysqli_stmt_store_result($stmt1);

        
// Check # of Records Returned.
        
if (mysqli_stmt_num_rows($stmt1)!==1){
            
// Friend-Request Count Found.

            // Bind result-set to variables.
            
mysqli_stmt_bind_result($stmt1$newFriendRequestsCount);

            
// Fetch record.
            
mysqli_stmt_fetch($stmt1);

            return 
$newFriendRequestsCount;

        }else{
            
// Friend-Request Count Not Found.
            
$resultsCode 'FUNCTION_FRIEND_REQUEST_COUNT_NOT_FOUND_5005';

            
// Set Source Page. (New)
            
$sourcePage $_SERVER['SCRIPT_NAME'];

            
// Log Function Error.
            
logError($dbc$resultsCode$sourcePage$requestee);

            
// End script.
//            exit();

// IS THIS OKAY TO DO, OR SHOULD I DO SOMETHING ELSE??
            
return '';
        }

    }
//End of getNewFriendRequestCount 

Originally, I had this line of code after my IF-THEN-ELSE...
PHP Code:
            return $newFriendRequestsCount

However, that caused an "Undefined Variable" error when the ELSE branch fired, so I moved things around, and just went with return '' for lack of a better idea?!

Suggestions?

Thanks,



Debbie