ช่วยหน่อยค่ะ...การแสดงข้อมูลในtextbox...งงมาหลายวันแล้วค่ะ

 
<?
if(isset($_POST['login']))
{
	if($_POST['user'] != "" && &POST['pass'] != "")
	{
		$sql = "SELECT * FROM member WHERE member.username = ".$_POST['user'];
		$obj = mysql_query($sql);
		while($result = mysql_fetch_array($obj))
		{
			if($_POST['user'] == $result['username'] && $_POST['pass'] == $result['password'])
			{
				header("Location : memberinfo2.php?id=".$result['id']);
			}
			else
			{
				header("Location : login.php");
			}
		}
	}
}
else
{
?>
<form method="post" name="formlogin">
<input type="text" name="user">
<input type="text" name="pass">
<input type="subbmit" name="login" value="Login">
</form>
<?
}
?>

<?
	$sql = "SELECT * FROM member WHERE member.id = ".$_GET['id'];  // $_GET['id'] ส่งผ่าน url มา
	$obj = mysql_query($sql);
	while($objresult = mysql_fetch_array($obj))
	{
		echo "<input type='text' name='user' value='$objresult['username']'><br>";
		echo "<input type='text' name='names' value='$objresult['name']'><br>";
		echo "<input type='text' name='lastnames' value='$objresult['lastname']'><br>";
		echo "<input type='text' name='tel' value='$objresult['tel']'><br>";
		echo "<input type='text' name='email' value='$objresult['email']'><br>";
	}
?>