Question

所以我是面向对象编程领域的新手，我目前正面临这个问题（代码中描述了所有内容）：

<?php
    class MyClass {
        // Nothing important here
    }

    class MyAnotherClass {
        protected $className;

        public function __construct($className){
            $this->className = $className;
        }
        public function problematicFunction({$this->className} $object){
            // So, here I obligatorily want an $object of
            // dynamic type/class "$this->className"
            // but it don't works like this...
        }
    }

    $object = new MyClass;
    $another_object = new MyAnotherClass('MyClass');

    $another_object->problematicFunction($object);
?>

任何人都可以帮助我吗？

谢谢，Maxime（来自法国：对不起我的英语）

Answer 1

您需要的是

public function problematicFunction($object) {
    if ($object instanceof $this->className) {
        // Do your stuff
    } else {
        throw new InvalidArgumentException("YOur error Message");
    }
}

Answer 2

试试这个

class MyClass {
    // Nothing important here
    public function test(){
        echo 'Test MyClass';
    }
}

class MyAnotherClass {
    protected $className;

    public function __construct($className){
        $this->className = $className;
    }
    public function problematicFunction($object){
        if($object instanceof $this->className)
        {
            $object->test();
        }
    }
}

$object = new MyClass;
$another_object = new MyAnotherClass('MyClass');

$another_object->problematicFunction($object);

Answer 3

这称为type hinting，你不想支持你想做的事。

如果所有这些动态类名都有共同之处（例如，它们是某些特性的不同实现），您可能希望定义一个基类（可能是抽象）类或接口，并将该共同祖先用作类型提示：< / p>

<?php

interface iDatabase{
    public function __contruct($url, $username, $password);
    public function execute($sql, $params);
    public function close();
}

class MyClass implements iDatabase{
    public function __contruct($url, $username, $password){
    }

    public function execute($sql, $params){
    }

    public function close(){
    }
}

class MyAnotherClass {
    protected $className;

    public function __construct($className){
        $this->className = $className;
    }
    public function problematicFunction(iDatabase $object){
    }
}

否则，只需将支票移至problematicFunction()正文内，其他答案即可解释。

PHP：函数参数必须是具有动态类名的Object

3 个答案: