I git this warning>>
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files (x86)\EasyPHP 2.0b1\www\HotelCalifornia\sites\reserve_room.php on line 85
Unknown column '' in 'where clause
This is my code>>
<?php
$user=$_GET['username'];
$Custno=$_GET['Custno'];
$RoomNo=$_GET['RoomNo'];
$username="jasmine";
$password="price";
$db=mysql_connect("localhost" , $username, $password);
mysql_select_db("hotel_cali", $db);
$sql="SELECT Email FROM customer WHERE username=\"$user\"";
$result=mysql_query($sql);
$myrow=mysql_fetch_array($result) or die(mysql_error());;
?>
How can i solve that?