连接链接不是资源?为什么?

我使用一个类来创建mysql连接并将链接存储在一个私有变量中，

但是当我在mysql_real_escape_string()中使用链接变量或尝试使用

时

is_resource()表示链接不是资源....这是我的类代码从library.php…

class db_system
{
    private $link;
    private $result;
    /*
        Constructor connects to database and store in link
    */
public function __construct()
{
    $lines = file('../../secret/topsecret');
    $this->link = mysqli_connect(trim($lines[0]), trim($lines[1]), trim($lines[2]), trim($lines[3])) or die("Could not connect to " . mysqli_error($this->link));
    if(is_resource($this->link))
        echo "true";
    else
        echo "false";  // it always prints false;
}
/*
    this method process the query and returns the result od the query
*/
public function sqlquery($query)
{
    $this->result = mysqli_query($this->link, $query) or die('query failed ' . mysqli_error($this->link));
}
/*
    this method returns the rows in an array form
*/
public function get_data()
{
    return mysqli_fetch_array($this->result);
}
/*
    destructor closes sql connection if link is active
*/
public function __destruct()
{
    if($this->link)
    {
        mysqli_close($this->link);
    }
}
public function getlink()
{
    return $this->link;
}
}

下面是view.php 中的代码

include 'library.php';
    $view = new db_system();  // connection object made
$per = "%";
$qry = sprintf('SELECT * FROM inventory WHERE description LIKE     "%s%s%s"',$per,mysql_real_escape_string($find,$view->getlink()),$per);

//$find是一个要搜索的值，它来自$_GET['find']，存储在$find中，不为空

之后，它说2参数不是mysql_real_escape_string().....中的资源

那么我如何使$link作为资源变量，以便我可以使用mysql_real_escape_string()

mysqli mysqli_connect ([ string $host = ini_get("mysqli.default_host") [, string $username = ini_get("mysqli.default_user") [, string $passwd = ini_get("mysqli.default_pw") [, string $dbname = "" [, int $port = ini_get("mysqli.default_port") [, string $socket = ini_get("mysqli.default_socket") ]]]]]] )