抽象静态方法中的实例子类

Question

我对 php 很陌生，我想从抽象父类的静态方法中实例化一个子类。是否可以在 php 中使用？

这是我的工作：

父类：

<?php

namespace App\views;


abstract class Views
{
    protected $template_name = '';

    abstract protected function Get();

    public static function as_view($target = null, $params = null) {
        // Here is this trick
        $create = function () {
            $cls = get_called_class();
            return new $$cls();
        };
        $instance = $create();
        $instance->Get();
    }

    public static function get_template() {
        return self::$template_name;
    }

}

儿童班：

<?php

namespace App\views;

use App\views\Views;

class HomeViews extends Views {
    protected $template_name = 'home.html';

    protected function Get() {
        echo self::get_template();
        dump($this);
    }
}

我想这样使用它：

<?php

use App\views\HomeView;

HomeView::as_view();

但是 php 抛出错误：

[Thu Sep 24 16:23:50 2020] 127.0.0.1:57001 [200]: / - Uncaught Error: Class name must be a valid object or a string in /Users/dev/PhpstormProjects/my_project/devel/src/views/Views.php:17
Stack trace:
#0 /Users/dev/PhpstormProjects/my_project/devel/src/views/Views.php(19): App\views\Views::App\views\{closure}()
#1 /Users/dev/PhpstormProjects/my_project/devel/public/index.php(9): App\views\Views::as_view()
#2 {main}

Answer 1

首先$template_name 应该是static，如果你想从静态方法中使用它。

您可以使用late static bindings 访问父类中的静态方法和字段：

<?php

abstract class Views
{
    protected static $template_name = '';

    abstract protected function Get();

    public static function as_view($target = null, $params = null) {
       $instance = new static();
       $instance->Get();
    }

    public static function get_template() {
        return static::$template_name;
    }

}
class HomeViews extends Views {
    protected static $template_name = 'home.html';

    protected function Get() {
        echo self::get_template() . "\n";
        var_dump($this);
    }
}

HomeViews::as_view();
/* outputs:
home.html
object(HomeViews)#1 (0) {
}
*/