我可以实例化另一个类中的PHP类吗？ [英] Can I instantiate a PHP class inside another class?

问题描述

我想知道是否允许在另一个类中创建一个类的实例。

I was wondering if it is allowed to create an instance of a class inside another class.

或者，我必须在外部创建它，然后传递通过构造函数？

Or, do I have to create it outside and then pass it in through the constructor? But then I would have created it without knowing if I would need it.

示例（数据库类）：

class some{

if(.....){
include SITE_ROOT . 'applicatie/' . 'db.class.php';
$db=new db

推荐答案

在另一个类中定义一个类。您应该包括类以外的其他类的文件。在你的情况下，这将给你两个顶级类 db 和一些。现在在一些的构造函数中，您可以决定创建一个 db 的实例。例如：

You can't define a class in another class. You should include files with other classes outside of the class. In your case, that will give you two top-level classes db and some. Now in the constructor of some you can decide to create an instance of db. For example:

include SITE_ROOT . 'applicatie/' . 'db.class.php';

class some {

    public __construct() {
        if (...) {
            $this->db = new db;
        }
    }

}

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