PHP 传入 $this 以在类外运行 [英] PHP pass in $this to function outside class

问题描述

您能否像在 javascript 中一样传入 $this 变量以在全局"空间中的函数中使用?我知道 $this 用于课程，但只是想知道.我试图避免使用全局"关键字.

Can you pass in a $this variable to use in a function in the "global" space like you can in javascript? I know $this is meant for classes, but just wondering. I'm trying to avoid using the "global" keyword.

例如:

class Example{
  function __construct(){ }
  function test(){ echo 'something'; }
}

function outside(){ var_dump($this); $this->test(); }

$example = new Example();

call_user_func($example, 'outside', array('of parameters')); //Where I'm passing in an object to be used as $this for the function

在 javascript 中，我可以使用 call 方法并分配一个 this 变量用于函数.只是好奇是否可以用 PHP 完成同样的事情.

In javascript I can use the call method and assign a this variable to be used for a function. Was just curious if the same sort of thing can be accomplished with PHP.

推荐答案

PHP 与 JavaScript 有很大不同.JS 是一种基于原型的语言，而 PHP 是一种面向对象的语言.在类方法中注入不同的 $this 在 PHP 中没有意义.

PHP is very much different from JavaScript. JS is a prototype based language whereas PHP is an object oriented one. Injecting a different $this in a class method doesn't make sense in PHP.

您可能正在寻找的是将不同的 $this 注入到闭包(匿名函数)中.这将可以使用即将推出的 PHP 5.4 版实现.请参阅对象扩展 RFC.

What you may be looking for is injecting a different $this into a closure (anonymous function). This will be possible using the upcoming PHP version 5.4. See the object extension RFC.

(顺便说一句，您确实可以将 $this 注入到不是 instanceof self 的类中.但正如我已经说过的，这没有任何意义完全没有.)

(By the way you can indeed inject a $this into a class which is not instanceof self. But as I already said, this doesn't make no sense at all.)

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