ใครเคยเจอปัญหาแบบนี้บ้างครับ คือว่าผมส่งสัยครับว่า สมมติว่าผมมีรูปอยู่แล้วแต่แค่ต้องการจะเปลี่ยนรูป

<? 
$fileupload=$_FILES['fileupload']['tmp_name'];
$fileupload_name=$_FILES['fileupload']['name'];
$fileupload_size=$_FILES['fileupload']['size'];
$fileupload_type=$_FILES['fileupload']['type'];

require"config/config_account.php";
if ($fileupload) {

	$array_last=explode(".",$fileupload_name);
	$c=count($array_last)-1; 
	$lastname=strtolower($array_last[$c]) ;
	
	if ($lastname=="gif" or $lastname=="jpg" or $lastname=="jpeg" or $lastname=="png") {

		$sql2="select * from $tb where Username='$Username' ";
		$result2=mysql_db_query($db,$sql2);
		$row=mysql_fetch_row($result2);

		$photoname=$row[0].".".$lastname;
		copy($fileupload,"picad/".$photoname);
		$sql3="UPDATE account SET photo_prd='$photoname' Where Username='$Username' ";
		$result3=mysql_db_query($db,$sql3);

	} 
	unlink($fileupload);

}

echo "<H3></H3>";

	




?>
<?
$Username=$_POST['Username'];
$Password=$_POST['Password'];
$Password2=$_POST['Password2'];
$Name=$_POST['Name'];
$Email=$_POST['Email'];
echo"$Username";
echo"$Password";
echo"$Name";
echo"$Email";
require"config/config_account.php";

$sql="select * from $tb WHERE Username='$Username'";
echo"$sql";
$db_query=mysql_db_query($db,$sql);
mysql_query("UPDATE $tb SET Password='$Password',Name='$Name',Email='$Email'
WHERE Username='$Username'");
mysql_close();

?>
<center><br><? include"connect_pic_ad.php";?><br><font color="#006600" size="2"><b>แก้ไขแล้ว</b></font><br><br>
<input type=button value="ปิดหน้าต่างนี้" onClick="javascript:window.close();"></center>
</body>
</html>