PHP如何从构造函数返回子类 [英] How to return subclass from constructor in PHP

问题描述

<?php

class Super {
    public $my;
    public function __construct ( $someArg ) {
        if ( class_exists('Sub') ) {    // or some other condition
            return new Sub( $someArg );
        }
        $this->my = $someArg;
    }
}

class Sub extends Super {}

?>

这不起作用，因为 new Super() 将是一个空"的 Super 对象(所有成员都是 NULL).(PHP 不允许对 $this 赋值，所以 $this = new Sub() 也不起作用).

This doesn't work, as new Super() will be an "empty" Super object (all members are NULL). (PHP doesn't allow assignments to $this, so $this = new Sub() doesn't work either).

我知道正确的模式应该是这里的工厂.但这需要对代码进行大量更改，所以我想知道是否可以这样做.由于 Sub 是一个 Super，我不明白为什么它不应该从 OOP 的角度受到限制.

I know the correct pattern would be a factory here. But that would require a lot of changes in the code, so I'm wondering whether it's possible to do it this way. Since Sub is-a Super, I don't see why it shouldn't be restricted from an OOP point of view.

推荐答案

您不能分配给 $this 并且不能从构造函数返回任何内容.

You can't assign to $this and you cannot return anything from a constructor.

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