cvfitmwhrx
New Member
This is the code I'm using to pass the database handle to my class:\[code\]$db = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);if ($db->connect_error) die('MySQL connection error (' . $db->connect_errno . ') ' . $db->connect_error);$linkgen = new Linkgen($db);echo $linkgen->getLinksLeft();\[/code\]And this is the code in the \[code\]Linkgen\[/code\] class:\[code\]class Linkgen { private $db; // our database handle private $linksLeft; function __contruct($db) { $this->db = $db; } function getLinksleft() { $ip = $_SERVER['REMOTE_ADDR']; $q = $this->db->prepare("SELECT links_left FROM `limits` WHERE ip=?"); $q->bind_param("s", $ip); $q->execute(); $q->bind_result($this->linksLeft); $q->fetch(); printf("%i links left", $this->linksLeft); $q->close(); }}\[/code\]For some reason I get the non-object error when I call the \[code\]getLinksLeft\[/code\] method. I can't figure this out because from what I can see I'm referencing the database handle correctly.Any help is appreciated, thanks.