查看数据库中联接表中的数据

即时通讯试图从2个单独的表中查看数据，但出现此错误：

"Notice: Trying to get property of non-object in D:\xampp\htdocs\testsubject\User\inventory.php on line 18"

这是我的PHP代码：

$sql = "SELECT storage_details.itemCODE,storage_details.pckgeID,storage_details.cndition,storage_details.duration,pckge_info.price,storage_details.status 
    FROM storage_details
    INNER JOIN pckge_info
    ON storage_details.pckgeID=pckge_info.pckgeID";
$result = $link->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo "itemcode: " . $row["itemCODE"]. " - packageid: " . $row["pckgeID"]. "condition: " . $row["cndition"]. "duration: " . $row["duration"]. " status: " . $row["price"]. " " . $row["status"]."<br>";
}
} else {
echo "0 results";
}

mysql_free_result($result);