Warning: mysql_fetch_array() expects parameter 1 to be ...

baiumbg

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Получавам следния error:

Код:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampplite\htdocs\login.php on line 14

В този код:

Код:
<?php
include("config.php");
if(!isset($_POST['submit'])) {
	echo '<form action="" method=post>
	Username: <input type=text name=username><br>
	Password: <input type=password name=password><br><br>
	<input type=submit name=submit value="Submit">';
}else{
	$conn=mysql_connect($dbhost, $dbuser, $dbpass);
	mysql_select_db($dbname, $conn) or die(mysql_error());
	$username=$_POST['username'];
	$query="SELECT FROM users WHERE username='$username'";
	$query1=mysql_query($query);
	while($row=mysql_fetch_array($query1)) {
		$pass=$row['password'];
		if($pass != $_POST['password']) {
			echo "Wrong Password! Click <a href='javascript:history.back()'>here</a> to go back<br>";
		}else{
			header('location: users.php');
		}
	}
}
?>

Отдавна не съм ползвал PHP :?
 
<?php
include("config.php");
if(!isset($_POST['submit'])) {
echo '<form action="" method=post>
Username: <input type=text name=username><br>
Password: <input type=password name=password><br><br>
<input type=submit name=submit value="Submit">';
}else{
$conn=mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname, $conn) or die(mysql_error());
$username=$_POST['username'];
$query="SELECT * FROM users WHERE username='$username'";
$query1=mysql_query($query);
while($row=mysql_fetch_array($query1)) {
$pass=$row['password'];
if($pass != $_POST['password']) {
echo "Wrong Password! Click <a href='javascript:history.back()'>here</a> to go back<br>";
}else{
header('location: users.php');
}
}
}
?>
 

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