PHP神奇常量-自动传递给函数？

我编写了一个简单的访问控制系统，它读取访问字符串数组，并根据结果返回true或false

我会这样称呼它（例如在类

user

的方法

list\u user\u data

中）：`

在里面，它检查当前用户是否有权限访问class

user

中的方法

list\u user\u data

它是有效的，但我发现我总是必须在调用中指定

\uuu类\uuuuuu

和

\uuuu方法\uuuuuu

，这让我很恼火。有没有一种方法可以从

current_user_can

函数中获取调用函数的值，这样我就可以简单地调用

current_user_can（）

，而不必传递神奇常数

我的代码按原样工作，但我认为它可以改进

这可能吗？

的返回值应该在第二个条目（索引1）中返回调用函数，例如：

<?php

function current_user_can()
{
    $backtrace = debug_backtrace(false, 2);
    // ToDo: Check if $backtrace[1] (and especially the class-key of that) actually exist...
    //       It should always (although the class key might not if this function isn't called from within a class), since this function is being called
    //       but it's still a good habbit to check array keys before accessing the array
    $callerClass = $backtrace[1]["class"];
    $callerMethod = $backtrace[1]["function"];

    // ToDo: implementation of check, in this example $callerClass would be "User" and $callerMethod would be "list_user_data"

    return true;
}

class User {
    public function list_user_data() {
        if (current_user_can())
        {

        }
    }
}

$user = new User();
$user->list_user_data();

---

也许值得一看。啊，这很有效，谢谢。请张贴作为我接受的答案。
<?php

function current_user_can()
{
    $backtrace = debug_backtrace(false, 2);
    // ToDo: Check if $backtrace[1] (and especially the class-key of that) actually exist...
    //       It should always (although the class key might not if this function isn't called from within a class), since this function is being called
    //       but it's still a good habbit to check array keys before accessing the array
    $callerClass = $backtrace[1]["class"];
    $callerMethod = $backtrace[1]["function"];

    // ToDo: implementation of check, in this example $callerClass would be "User" and $callerMethod would be "list_user_data"

    return true;
}

class User {
    public function list_user_data() {
        if (current_user_can())
        {

        }
    }
}

$user = new User();
$user->list_user_data();