Ik krijg deze fout, maar ik snap het niet.
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>";
	}
?>
Het is nu wel duidelijk.
Ik doe dit topic maar op slot.

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