simple login form using php and mysql code example
Example 1: simple login form in php
<?php
session_start();
$errorMsg = "";
$validUser = $_SESSION["login"] === true;
if(isset($_POST["sub"])) {
$validUser = $_POST["username"] == "admin" && $_POST["password"] == "password";
if(!$validUser) $errorMsg = "Invalid username or password.";
else $_SESSION["login"] = true;
}
if($validUser) {
header("Location: /login-success.php"); die();
}
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="content-type" content="text/html;charset=utf-8" />
<title>Login</title>
</head>
<body>
<form name="input" action="" method="post">
<label for="username">Username:</label><input type="text" value="<?= $_POST["username"] ?>" id="username" name="username" />
<label for="password">Password:</label><input type="password" value="" id="password" name="password" />
<div class="error"><?= $errorMsg ?></div>
<input type="submit" value="Home" name="sub" />
</form>
</body>
</html>
Example 2: Login HTML PHP
<?php
session_start();
$pdo = new PDO('mysql:host=localhost;dbname=test', 'root', '');
if(isset($_GET['login'])) {
$email = $_POST['email'];
$passwort = $_POST['passwort'];
$statement = $pdo->prepare("SELECT * FROM users WHERE email = :email");
$result = $statement->execute(array('email' => $email));
$user = $statement->fetch();
//Überprüfung des Passworts
if ($user !== false && password_verify($passwort, $user['passwort'])) {
$_SESSION['userid'] = $user['id'];
die('Login erfolgreich. Weiter zu <a href="geheim.php">internen Bereich</a>');
} else {
$errorMessage = "E-Mail oder Passwort war ungültig<br>";
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Login</title>
</head>
<body>
<?php
if(isset($errorMessage)) {
echo $errorMessage;
}
?>
<form action="?login=1" method="post">
E-Mail:<br>
<input type="email" size="40" maxlength="250" name="email"><br><br>
Dein Passwort:<br>
<input type="password" size="40" maxlength="250" name="passwort"><br>
<input type="submit" value="Abschicken">
</form>
</body>
</html>