调用类方法（使用构造函数）在php中没有对象实例化

本文介绍了调用类方法（使用构造函数）在php中没有对象实例化的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

Ive看了看，试过，但我找不到答案。

Ive looked and tried but I can't find an answer.

在PHP中，是否可以调用类的成员函数（当该类需要一个构造函数来接收参数时）而不将其实例化为一个对象？

In PHP, is it possible to call a class' member function (when that class requires a constructor to receive parameters) without instantiating it as an object?

代码示例（给出错误）：

A code example (which gives errors):

<?php class Test { private $end=""; function __construct($value) { $this->end=$value; } public function alert($value) { echo $value." ".$this->end; } } //this works: $example=new Test("world"); $example->alert("hello"); //this does not work: echo Test("world")::alert("hello"); ?>

推荐答案

不幸的是PHP不支持这样做，但你是一个有创意的人：D

Unfortunately PHP doesn't have support to do this, but you are a creative and look guy :D

你可以使用工厂，样例：

You can use an "factory", sample:

<?php class Foo { private $__aaa = null; public function __construct($aaa) { $this->__aaa = $aaa; } public static function factory($aaa) { return new Foo($aaa); } public function doX() { return $this->__aaa * 2; } } Foo::factory(10)->doX(); // outputs 20