HELP! needed with PHP login script linked to MySQL database
Posted: Fri Aug 19, 2011 3:30 pm
Hey peeps can anyone advise me on the problem that I have coming up with one of my PHP login scripts that is used to connect to a MySQL database for login verification. I'm pretty new to PHP and MySQL so sorry for asking for help with for something that is probably going to turn out to be something so silly but i really am stuck.
In the following main code:
Also I have the following external script associated with the above:
But when run im getting the following message displayed in the browser:
Any ideas??????
In the following main code:
Code: Select all
<html>
<body>
<form action="index3.php?login=yes" method="POST">
Username:<input type="text" name="user"><br />
Password:<input type="password" name="pass" ><br/>
<input type="submit" name="login" value="SIGN-IN" ><p>
</form>
<?php
$user=$_POST['user'];
$pass=$_POST['pass'];
$login=$_POST['login'];
echo $login;
function denied()
{
echo '<h3><span style= "color:red"> Access Denied!!! </span></h3><br><br>';
}
function granted ($user)
{
echo '<h3><span style= "color:green"> Access Granted!!! </span></h3>';
echo 'Welcome, ' . $user;
}
if($login=='SIGN-IN')
{
$con= include_once "mysql_connect.php";
$get = mysql_query("SELECT count(id) FROM login WHERE user='$user' and pass='$pass'");
$result = mysql_result($get, 0);
if (empty($user) || empty($pass))
{
//stop execution of PHP script using die function!
denied();
die("<br>Please fill out user login fields carefully....<br>");
}
if ($result!=1)
{
granted($user);
}
else
{
denied ();
//$_SESSION['user']=$user;
}
}
?>
</body>
</html>
Code: Select all
<?php
$db_host = "localhost";
$db_username = "root";
$db_pass = "rhianna";
$db_name = "signin";
@mysql_connect("$db_host", "$db_username", "$db_pass") or die ("Could not connect to MySQL");
@mysql_select_db("$db_name") or die ("No $db_name Database ");
?>
I Think the problem lies with in mysql_result() function of the main script when using the $get variable as one of its parameters, but have absolutely no idea what it could be or how it can be fixed:Warning: mysql_result() expects parameter 1 to be resource, string given in C:\xampp\htdocs\index3.php on line 44
Code: Select all
$get = mysql_query("SELECT count(id) FROM login WHERE user='$user' and pass='$pass'");
$result = mysql_result($get, 0);