Question

Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

I have next class:

<?php
    class DbHelper{
        private $databaseURL;
        private $databaseUName;
        private $databasePWord;
        private $databaseName;
        private $nameOfDbWithWorkers;
        private $connection;

        function __construct($dbURL, $dbUserName, $dbPword, $dbName, $nameOfDbWithWorkers){
            $this->databaseURL = $dbURL;
            $this->databaseUName = $dbUserName;
            $this->databasePWord = $dbPword;
            $this->databaseName = $dbName;
            $this->nameOfDbWithWorkers = $nameOfDbWithWorkers;
        }

        function setConnectionToDb(){
            $this->connection = mysql_connect($this->databaseURL,$this->databaseUName,$this->databasePWord) OR DIE("can't connect to DB"); 
            mysql_select_db($this->databaseName, $this->connection)or die ("Error while connecting to database");
        }

        function getUser($login, $pass){
            echo "$login, $pass";
            $query = "SELECT type FROM $this->nameOfDbWithWorkers WHERE login = '$login' and password = '$pass';"; 
            $queryResult = $this->getDataFromDbByQuery($query);
            if ((mysql_affected_rows($this->connection) == 1)){
                $meta = mysql_fetch_assoc($queryResult);
                if ($meta['type']=='admin'){
                    return 'admin';
                }
                if ($meta['type']=='user'){
                    return 'user';
                }
                else{
                    return 'nomatch';
                }
            }
            else{               
                 return 'nomatch';
            }           
        }

        function getDataFromDbByQuery($query){
            $this->setConnectionToDb();
            $result = mysql_query($query);
            mysql_close($this->connection);
            return $result;
        }
}
?>

and invoke this by

$dbHelp = new DbHelper($dbURL, $dbUName, $dbPword, $dbName, $nameOfDbWithWorkers);
$userType = $dbHelp->getUser($login, $pass);

And have a next error:

Warning: mysql_affected_rows(): 5 is not a valid MySQL-Link resource in Z:\home\ecl.ru\www\classes\dbhelper.php on line 27

What’s wrong?

Answer 1

first , use mysql_num_rows for row count , mysql_affected_rows returns number of changed rows , as is written in manual on php.net

Get the number of affected rows by the last INSERT, UPDATE, REPLACE or DELETE query associated with link_identifier.

second – evry time you are using mysql function , put there link identifier or result identifier