变量未传递给类方法

Question

我正在使用PHP构建一个登录类,但是当我将它们传递给类方法时使用的变量是空的,即使它们不应该.我尝试只返回$ username变量,但它仍然是空的,但如果我不使用类返回它,我可以看到它被正确分配.

我正在使用多个其他类以及正确分配变量的方法.

我不知道自己是否瞎了眼,错过了一些显而易见的东西,或者是否还有其他因素造成的.

class Auth
{
    private $mysqli;

    public function __construct(mysqli $mysqli)
    {
        $this->mysqli = $mysqli;
    }

    public function login($username, $password) //These variables are empty, even when they shouldn't be
    {
        $return['error'] = true;

        $uid = $this->getUserId(strtolower($username)); //Returns false because $username variable is empty

        if (!$uid) {
            $return['message'] = 'No such user.'; //Output
            return $return;
        }

        $user = $this->getUser($uid);

        if (!password_verify($password, $user['password'])) {
            $return['message'] = 'Password incorrect';
            return $return;
        }

        $return['error'] = false;
        $return['message'] = 'Logged in';

        return $return;
    }

    private function getUserId($username)
    {
        $stmt = $this->mysqli->prepare("SELECT id FROM users WHERE username = ?");
        $stmt->bind_param('s', $username);
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($id);

        if ($stmt->num_rows < 1) {
            return false;
        }

        $stmt->fetch();

        return $id;
    }

    private function getUser($uid)
    {
        $stmt = $this->mysqli->prepare("SELECT username, password, email FROM users WHERE id = ?");
        $stmt->bind_param('s', $uid);
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($username, $password, $email);

        if ($stmt->num_rows < 1) {
            return false;
        }

        $stmt->fetch();

        $return['uid'] = $uid;
        $return['username'] = $username;
        $return['password'] = $password;
        $return['email'] = $email;

        return $return;
    }
}

表单分配发送的变量.

<form method="POST" action="post.php">
    <label>Username
    <input style="display:block;width:250px;" type="text" name="username" required></label>
    <label>Password
    <input style="display:block;width:250px;" type="password" name="password"></label>
    <button style="display:block;" class="default_btn">Log in</button>
</form>

if (isset($_POST['username'])) {
    $username = $_POST['username'];
    $password = $_POST['password'];

    $auth = new Auth($mysqli);
    $auth->login($username, $password);

    if ($auth->login()['error']) {
        echo 'error:' . $auth->login()['message'];
    } else {
        echo 'success:' . $auth->login()['message'];
    }
}

编辑:

如果我在类方法中分配变量,则代码可以工作:

public function login($username = 'user', $password = 'pass')

但如果我这样做,它将无法工作:

$username = 'User';
$password = 'pass';

$auth = new Auth($mysqli);
$auth->login($username, $password);

此外,如果我使用$ auth-> login()之外的$ _POST值,则会将它们分配,以便在将它们传递给类时它们不为空...

Answer 1

问题似乎是您不存储结果，而是在没有值的情况下再次调用登录：if ($auth->login()['error']) {。

尝试这个：

<form method="POST" action="post.php">
    <label>Username
    <input style="display:block;width:250px;" type="text" name="username" required></label>
    <label>Password
    <input style="display:block;width:250px;" type="password" name="password"></label>
    <button style="display:block;" class="default_btn">Log in</button>
</form>

if (isset($_POST['username'])) {
    $username = $_POST['username'];
    $password = $_POST['password'];

    $auth = new Auth($mysqli);
    $login_result = $auth->login($username, $password);

    if ($login_result['error']) {
        echo 'error:' . $login_result['message'];
    } else {
        echo 'success:' . $login_result['message'];
    }
}