search แล้วไม่ออกมาทำไงดี errorตลอด แก้ไม่เป็นช่วยผมทีครับ

$dbname = "SELECT * FROM teacher WHERE (name LIKE '%".$_GET["search"]."%')";

$dbname = "SELECT * FROM teacher WHERE (name LIKE '".$_GET["search"]."')";
$passwd = mysql_query($dbname) or die ("Error Query [".$dbname."]");
$sql=mysql_db_query($dbname, $sql);  // บรรทัดนี้มีเพื่อ.???

while ($array = mysql_fetch_array($dbname)) // น่าจะ  $array = mysql_fetch_array($passwd)

<?php

if($_GET["search"] != "")
{ 
// Search By Name or phone
include ("config.php");
$sql = "SELECT * FROM teacher WHERE (name LIKE '%".$_GET["search"]."%')";
$query = mysql_query($sql);

?>
<table width="600" border="1">
<tr>
<th width="91"> <div align="center">Name </div></th>
<th width="98"> <div align="center">Level </div></th>
<th width="198"> <div align="center">Phone </div></th>
</tr>

<?
while ($array = mysql_fetch_array($query))
{
?>
<tr>
<td><div align="center"><?=$array["name"];?></div></td>
<td><?=$array["level"];?></td>
<td><?=$array["phone"];?></td>
</tr>
<?
}
?>
</table>
<?
mysql_close();
}
?>

<?php

if($_GET["search"] != "")
  {  
    // Search By Name or phone
include ("config.php");
    $dbname = "SELECT * FROM teacher WHERE (name LIKE '%".$_GET["search"]."%' or phone LIKE '%".$_GET["search"]."%' )";
    $passwd = mysql_query($dbname) or die ("Error Query [".$dbname."]");
    ?>
    <table width="600" border="1">
      <tr>
        <th width="91"> <div align="center">Name </div></th>
        <th width="98"> <div align="center">Level </div></th>
        <th width="198"> <div align="center">Phone </div></th>
      </tr>
      
    <?
    while ($array = mysql_fetch_array($passwd)) ผมเป็นตัวนี้มาเป็น $passwd จากเดีม $dbname
    {
   ?>
     <tr>
       <td><div align="center"><?=$array["name"];?></div></td>
     <td><?=$array["level"];?></td>
     <td><?=$array["phone"];?></td>
     </tr>
   <?
    }
    ?>
    </table>
    <?
    mysql_close();
	}
?>