我正在使用mysql数据库创建一个登录系统。
数据库中有两个表，分别是student和discipline。我希望用户从不同的表中使用两种不同的东西登录。我希望他们使用student表中的student id和discipline表中的discipline name登录。学生和纪律都是分开的。我已经创建了一个与表的内部连接，查询在phpmyadmin中工作。但是，运行应用程序时无法登录。当我按“登录”按钮时，似乎什么都没发生。。。。
下面是我的代码：

// LOGIN USER
if (isset($_POST['login'])) {
    $name = mysqli_real_escape_string($db, $_POST['name']);
    $studentid = mysqli_real_escape_string($db, $_POST['studentid']);

}

    if (count($errors) == 0) {
        $name = md5($name);
        $query = "SELECT discipline.name, student.studentid 
        FROM discipline
        INNER JOIN student ON discipline.disciplineid = student.disciplineid WHERE discipline.name='$name' AND student.studentid='$studentid'";

        //$query = "SELECT * FROM discipline WHERE name='$name' AND studentid='$studentid'";

        $result = mysqli_query($db, $query);
    }
        if (mysqli_num_rows($result) == 1) {
            $_SESSION['name'] = $name;
            header('location: Assessment.php');
        }else {
            array_push($errors, "Wrong username/password combination");

        }

//logout
if (isset($_GET['logout'])) {
    session_destroy();
    unset($_SESSION['disciplineid']);
    header('location: index.php');
}
?>

表单html:

<div class="form">
    <ul class="row justify-content-center">
        <form method="post" action="index.php">
            <p> <label>Your ID </label> <input type="text" id="studentid" name="studentid" required /> </p>
            <p> <label>name of discipline:</label> <input type="text" id="name" name="name"required /> </p>
            <p> <button type="submit" class="btn" name="login">Login</button> </p>
        </form>

任何帮助都将不胜感激。谢谢