Als ik naar de code kijk, klopt het wel.
Maar toch zeurt hij om een fout.
Parse error: syntax error, unexpected variable "$pass"
<?php
if($_POST['submit'] {
$user = $_POST['username']
$pass = md5($_POST['password'])
$key = $_POST['key']
$sql = "SELECT * FROM AdminUsers WHERE Username='".$user."' AND Passwd='".$pass."' AND LoginKey='".$key."'";
$result = mysqli_query($connect, $sql);
$logindetails = mysqli_fetch_all($result, MYSQLI_ASSOC);
foreach($logindetails as $detail) {
$_SESSION['Name'] = $detail['Name'];
$_SESSION['Username'] = $detail['Username'];
$_SESSION['Passwd'] = $detail['Passwd'];
$_SESSION['Level'] = $detail['Access'];
$_SESSION['ID'] = $detail['ID'];
}
header('location: ./inc/start.php');
exit();
echo "</div>";
}else{
echo"
<div class='login'>
<form id='login' action='".htmlspecialchars($_SERVER["PHP_SELF"]."' method='POST'>
<fieldset>
<input type='text' name='username' placeholder='Username'>
<input type='password' name='password' placeholder='Password'>
<input type='text' name='key' placeholder='Keycode'>
<button type='submit' name='submit'>Login</button>
</fieldset>
</form>
<img id='img-login' src='./img/page/Sexy-Devil.jpg'>
</div>";
}
?>