php - 表单注册没有收到错误

Question

所以我编辑了我的代码但仍然无法获得结果部分代码负责报告不起作用的错误 我愿意接受其他编码方式来完成这项工作，只要它适合这里的代码

<?php

require 'core.inc.php';






if(!loggedIn()) {

//check mikunim ke tamame field ha dar form vojod darand va set shudan

if(isset($_POST['username'])&&isset($_POST['password'])&&isset($_POST['password_again'])&&isset($_POST['firstname'])&&isset($_POST['surname'])) {

    $username = $_POST['username'];
    $password = $_POST['password'];
    $password_again = $_POST['password_again'];
    $firsname = $POST['firstname'];
    $surename = $POST['surename'];



    //HALA CHECK MIKUNIM KHALI HASTAND YA NA

    if(!empty($username)&&!empty($password)&&!empty($email)&&!empty($firstname)&&!empty($surename)){
        echo 'ok' ;

    } else {

        echo ' All fields are required';

    }

}   

?>

<form action="register.php" method="POST">
Username:<br> <input type="text" name="username"><br> <br>
Password:<br> <input type="password" name="password"><br><br>
Password again:<br> <input type="password" name="password_again"><br><br>
Firstname:<br> <input type="text" name="firstname"><br><br>
Surname:<br> <input type="text" name="surename"><br><br>
<input type="submit" value="register">



</form>

<?php
    } else if (loggedIn()) {

    echo 'you \'re already logged in';
    }

我再次编辑，现在我明白了

注意:未定义索引:HTTP_REFERER

$http_referer = $_SERVER['HTTP_REFERER']

我在教程中使用了这个

Answer 1

改变这个

$firsname = $POST['firstname'];
$surename = $POST['surename'];

为此:

$firstname = $_POST['firstname'];
$surename = $_POST['surename'];