Hai guys. So my problem is that My user.inc.php file is all correct but it shows the error
[syntax=php]Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/a4495253/public_html/core/inc/user.inc.php on line 19[/syntax]
My user.inc.php file:
[syntax=php]<?php
// Checks if the given username exists in the table.
function user_exists($user){
$user = mysql_real_escape_string($user);
$total = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `user_name` = '{$user}'");
return (mysql_result($total, 0) == '1') ? true : false;
}
// Checks if the given username and password combination is valid.
function valid_credentials($user, $pass){
$user = mysql_real_escape_string(htmlentities($user));
$pass = sha1($pass);
mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `user_name` = '{$user}' AND `user_password` = '{$password}'");
return (mysql_result($total, 0) == '1') ? true : false;
}
// Adds a user to the Database.
function add_user($user, $pass, $first_name, $last_name){
$user = mysql_real_escape_string(htmlentities($user));
$pass = sha1($pass);
$first_name = mysql_real_escape_string(htmlentities($first_name));
$last_name = mysql_real_escape_string(htmlentities($last_name));
mysql_query("INSERT INTO `users` (`user_name`, `user_password`, `first_name`, `last_name`) VALUES ('{$user}', '{$pass}', '{$first_name}', '{$last_name}')");
}
?>[/syntax]
login.php:
[syntax=php]<?php
include('core/init.inc.php');
$errors = array();
if (isset($_POST['username'], $_POST['password'])){
if (empty($_POST['username'])){
$errors[] = 'Please enter your Username!';
}
if (empty($_POST['password'])){
$errors[] = 'Please enter your Password!';
}
if (valid_credentials($_POST['username'], $_POST['password']) === false){
$errors[] = 'Username or Password is incorrect!';
}
if (empty($errors)){
$_SESSION['username'] = htmlentities($_POST['username']);
header('Location: home.php');
die();
}
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Login to your Profile!</title>
<meta name="description" content="Welcome to Socialcrunch. You may connect with friends, meet new ones, and also hang out when you have free time!" />
<meta name="keywords" content="social, networking, games, chat, hangout, friends" />
<meta name="robots" content="index, follow" />
<link rel="stylesheet" href="styles/form.css" type="text/css">
<script type="text/javascript">
WebFontConfig = {
google: { families: [ 'Grand+Hotel::latin' ] }
};
(function() {
var wf = document.createElement('script');
wf.src = ('https:' == document.location.protocol ? 'https' : 'http') +
'://ajax.googleapis.com/ajax/libs/webfont/1/webfont.js';
wf.type = 'text/javascript';
wf.async = 'true';
var s = document.getElementsByTagName('script')[0];
s.parentNode.insertBefore(wf, s);
})(); </script>
</head>
<body>
<div id="wrapper">
<div class="login">
<div>
<?php
if (empty($errors) === false){
?>
<ul>
<?php
foreach ($errors as $error){
echo "<li>{$error}</li>";
}
?>
</ul>
<?php
}
?>
</div>
<center>
<b>Login to an Existing Profile.</b>
<form action="" method="post">
<p class="username">
Username: <input type="text" name="username" id="username" value="<?php if (isset($_POST['username'])) echo htmlentities($_POST['username']); ?>" /><br />
</p>
<p class="password">
Password: <input type="password" name="password" id="password" /><br />
</p>
<p class="submit">
<input type="submit" name="submit" value="Login"><br />
</p>
</form>
Don't have a Profile yet? <a href="signup.php">Sign up!</a><br />
Forgot your Password? <a href="">Reset Password</a>
</center>
</div>
</div>
</body>
</html>[/syntax]
Login Probs again D:
-
ExtremeGaming
- Posts: 205
- Joined: Mon Jul 09, 2012 11:13 pm
Re: Login Probs again D:
[syntax=php]
mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `user_name` = '{$user}' AND `user_password` = '{$password}'");
return (mysql_result($total, 0) == '1') ? true : false;
[/syntax]
You aren't giving your query the variable name of $total so mysql_result is throwing an error because there's no query to act upon
mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `user_name` = '{$user}' AND `user_password` = '{$password}'");
return (mysql_result($total, 0) == '1') ? true : false;
[/syntax]
You aren't giving your query the variable name of $total so mysql_result is throwing an error because there's no query to act upon
<?php while(!$succeed = try()); ?>
Re: Login Probs again D:
Just to clarify the above, you need to assign the result of the query to a variable so you can use it lower down.
[syntax=php]$total = mysql_query( ...[/syntax]
[syntax=php]$total = mysql_query( ...[/syntax]
