PHP：如何从属性“className”实例化 class

Question

想象一下这段代码：

class MyClass
{
    private string $className;
    public function __construct(string $className)
    {
        $this->className = $className;
    }

    public function instantiateClass()
    {
        $className = $this->className;
        return new $className();
    }
}

有没有办法实例化 class 而不首先在方法instantiateClass()中将属性值分配给局部变量$className ？

像这样的东西：

class MyClass
{
    private string $className;
    public function __construct(string $className)
    {
        $this->className = $className;
    }

    public function instantiateClass()
    {
        // This cannot be done as 'className' should be a method, not the property
        return new $this->className();
    }
}

有任何想法吗？

Answer 1

因此，正如@Cid 在我的问题下面的评论中指出的那样，解决方案实际上是我认为错误的解决方案：

class MyClass { 私有字符串 $className; 公共 function __construct(string $className) { $this->className = $className; }

public function instantiateClass()
{
    // This works! It doesn't find a method, but reads the property correctly!
    return new $this->className();
}

}

Answer 2

您可以使用__CLASS__或self::或static::

你可以试试

class MyClass
{

    public function __construct()
    {
    }

    static public function instantiateClass()
    {
        return new self();
    }

    static public function instantiateClassWithStatic()
    {
        return new static();
    }

}

$myInstance = MyClass::instantiateClass();