PHP header 认证奇怪代码

Question

我从 PHP.net 复制代码用于 header 身份验证

代码如下：

<?php
$valid_passwords = array ("Itay" => "1234", "beta" => "password");
$valid_users = array_keys($valid_passwords);
$user = $_SERVER['PHP_AUTH_USER'];
$pass = $_SERVER['PHP_AUTH_PW'];
$validated = (in_array($user, $valid_users)) && ($pass == $valid_passwords[$user]);
if (!$validated) {
  header('WWW-Authenticate: Basic realm="this is testing area. beta testers only!"');
  header('HTTP/1.0 401 Unauthorized');
  die ("Not authorized");
}
// If arrives here, is a valid user.
?>

现在，我的问题是为什么这一行：

$validated = (in_array($user, $valid_users)) && ($pass == $valid_passwords[$user]);

使用

代替

$validated = $pass == $valid_passwords[$user];

Answer 1

$validated = (in_array($user, $valid_users)) && ($pass == $valid_passwords[$user]);

此行检查用户是否存在以及密码是否正确。

如果去掉user校验，那么user="nonexisting"和password=""就有效了（当然会有提示。

检查：

$pass = "";
$user = "any non existing"
$validated = ($pass == $valid_passwords[$user]); //here will be notice
var_dump($validated);

PHP header 认证奇怪代码

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