I am unable to login even tho I provided correct username and password, this is error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak/index.php on line 79 (I am not sure what this error means, since I don't want to show any rows, I don't see purpose of this code in line 79
$result = mysql_fetch_array($sql);
ERROR: Neispravno korisnicko ime! (this error means I inserted wrong username even tho I did not)
and this is code from page, I have created database and tables, I deleted login from this code
if you need to try by yourself here is link Click Here and user administrator and password webservisi
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Login</title>
<style type="text/css">
<!--
#form1 table {
background-color: #003333;
}
.style1 {
font-family: "Times New Roman", Times, serif;
font-size: 14px;
font-style: normal;
color: #FFFFFF;
}
.style2 {
font-family: "Times New Roman", Times, serif;
font-size: 18px;
font-style: normal;
color: #FFFFFF;
}
body {
background-color: #FFFFFF;
}
-->
</style>
</head>
<body>
<form id="form1" name="form1" method="post" action="index.php">
<p> </p>
<p> </p>
<table width="100%" height="308" align="center" bordercolor="#000000">
<tr>
<td colspan="2"><div align="center"><img src="Slike/Logo it.JPG" width="262" height="76" /></div></td>
</tr>
<tr>
<td width="426"> </td>
<td width="422"> </td>
</tr>
<tr>
<td><div align="right" class="style1">
<p class="style1"><span class="style1">Korisnicko ime</span>:</p>
</div></td>
<td><input type="text" name="username" id="username" /></td>
</tr>
<tr>
<td><div align="right" class="style1">Lozinka: </div>
<label></label></td>
<td><input type="password" name="password" id="password" /></td>
</tr>
<tr>
<td colspan="2"><label>
<label></label>
<?php
//Podaci o bazi podataka
$hostname = "";
$db_username = "";
$db_password = "";
$db_name = "";
if (isset($_POST['login'])){
//Provjera korisnika
$username = $_POST['username'];
$password = $_POST['password'];
if (empty($username) || empty($password)){
die('<hr>ERROR: MORATE POPUNITI SVA POLJA. <a
href="index.php">LogIn</a>.<br>');
}
$con = mysql_connect($hostname,$db_username,$db_password);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db_name, $con);
$sql = mysql_query("select password from students where username ='".$username."'");
$result = mysql_fetch_array($sql);
if(!$result)
die ("ERROR: Neispravno korisnicko ime!");
if ($password != $result['password']){
die ("ERROR: Neispravna lozinka!");
}
session_start();
$_SESSION['korisnik'] = $username;
if ($username == "administrator"){
header("Location: admin.php");
}else{
header("Location: user.php");
}
mysql_close($con);
}
?>
<div align="center">
<input type="submit" name="login" id="login" value="Prijava" />
</div>
</label></td>
</tr>
<tr>
<td colspan="2"><div align="center" class="style1">
<p class="style2">Provjera predmeta koje student može da sluša </p>
</div></td>
</tr>
</table>
<p> </p>
<p align="center">
<label></label>
</p>
<p align="center">
<label></label>
</p>
</form>
</body>
</html>
I tried changing $sql = mysql_query("select password from students where username ='".$username."'");
students into Students, because in database table is called Students, but then I get different type of error
thanks in advance