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PHP构造方法问题

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Release: 2016-06-06 20:39:35
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初学PHP构造方法,有这样一个简单的构造方法问题,请教如下:

<code>class Name{
    //成员属性
    public $name;
    //构造方法
    public function __construct($name = 'php'){
        $this->name = $name;
    }
}
</code>
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请解释一下:$this->name = $name; 这句怎么理解,是将$name的收到或者初始化的值赋给本类中的$name吗?还是怎么理解,非常感谢。

回复内容:

初学PHP构造方法,有这样一个简单的构造方法问题,请教如下:

<code>class Name{
    //成员属性
    public $name;
    //构造方法
    public function __construct($name = 'php'){
        $this->name = $name;
    }
}
</code>
Copy after login
Copy after login

请解释一下:$this->name = $name; 这句怎么理解,是将$name的收到或者初始化的值赋给本类中的$name吗?还是怎么理解,非常感谢。

是,Name::$name是类属性,实例出来的话就是对象属性。

new Name('php') 会创建一个 Name 的对象(或者叫做实例),将这个对象的 name 属性设置为 'php'

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