ช่วยดูโค้ดหน่อยคับ มัน Error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 <?
			 
include("config.php");
	  $strZ = "SELECT * FROM interview ORDER BY interview_id  ASC ";
$objQueryZ = mysql_db_query($dbname,$strZ);


$num=0;
while($objResultZ = mysql_fetch_array($objQueryZ)){ $num++;

//if(empty($objResultID["interview_year"] )){echo "**********************ยังไม่มีข้อมูล**********************";}else{
?>

           
       	<tr>
        	<td  ><div align="center"><? echo $objResultZ["interview_year"]?></div></td>
            <td  ><div align="center"><? echo $objResultZ["interview_year"]?></div></td>
            <td  ><div align="center"><? echo $objResultZ["interview_term"]?></div></td>
            <td  ><div align="center"><? echo $objResultZ["interview_open"]?></div></td>
            <td  ><div align="center"><? echo $objResultZ["interview_close"]?></div></td>
            <td  ><div align="center"><? echo $objResultZ["interview_day"]?></div></td>
            <td  ><div align="center"><? echo $objResultZ["interview_year"]?></div></td>
            <td  ><div align="center"><? echo $objResultZ["interview_year"]?></div></td>
        </tr>
      <? }//}?>