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Home Backend Development PHP Tutorial PHP类属性的权限问题 ctrl+v 第一次用第一次被喷 正视过后仅以警醒

PHP类属性的权限问题 ctrl+v 第一次用第一次被喷 正视过后仅以警醒

Jun 06, 2016 pm 08:48 PM
php

让我无法理解的代码,哪位老师或高手能指点指点,万分感谢。
代码如下:

<?php
class person {
    private $n1;
    private $n2;
    private $n3;

    public function __set($pro_name, $pro_val){
        $this->{$pro_name} = $pro_val;
    }

    public function __get($pro_name) {
        if (!isset($this->{$pro_name})) {
            return null;
        } else {
            return $this->{$pro_name};
        }
    }
}


class a {
    public $n;
    public $x;

    public function __construct($in, $ix) {
        $this->n = $in;
        $this->x = $ix;
    }
}

$pef1 = new person();
$pef1->n1 = '小明';
$pef1->n2 = 200;
echo $pef1->n1, '||', $pef1->n2;

echo PHP_EOL;

$c = new a('小小', 2003);
echo $c->n, '||', $c->x;
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Copy after login

第一个person类和第二个a类的调用问题,第一个类的成员属性是私有的,老师说过可以用__set__get来统一来管理,这个没问题。可问题是创建对象的时候,我可以不用和创建的对象名一致就能调用。比如说:我创建一个person类,然后再创建一个$p1=new person();对象,然后我调用,正常是这个的$p1->n1="小明";但现在我却不用这样也能调用,$uu->n1="小明";。这是为什么呢?

第二个a类主要是我为了证明是不是所有的类在创建对象后,都可以不用对象名来调用,结果发现不行。这个我可以理解,但第一种我就不能理解了。为什么protectedprivate的成员属性和方法在使用了__set__get这一对函数后在对象调用的时候就可以不用对象名呢?

回复内容:

让我无法理解的代码,哪位老师或高手能指点指点,万分感谢。
代码如下:

<?php
class person {
    private $n1;
    private $n2;
    private $n3;

    public function __set($pro_name, $pro_val){
        $this->{$pro_name} = $pro_val;
    }

    public function __get($pro_name) {
        if (!isset($this->{$pro_name})) {
            return null;
        } else {
            return $this->{$pro_name};
        }
    }
}


class a {
    public $n;
    public $x;

    public function __construct($in, $ix) {
        $this->n = $in;
        $this->x = $ix;
    }
}

$pef1 = new person();
$pef1->n1 = '小明';
$pef1->n2 = 200;
echo $pef1->n1, '||', $pef1->n2;

echo PHP_EOL;

$c = new a('小小', 2003);
echo $c->n, '||', $c->x;
Copy after login
Copy after login

第一个person类和第二个a类的调用问题,第一个类的成员属性是私有的,老师说过可以用__set__get来统一来管理,这个没问题。可问题是创建对象的时候,我可以不用和创建的对象名一致就能调用。比如说:我创建一个person类,然后再创建一个$p1=new person();对象,然后我调用,正常是这个的$p1->n1="小明";但现在我却不用这样也能调用,$uu->n1="小明";。这是为什么呢?

第二个a类主要是我为了证明是不是所有的类在创建对象后,都可以不用对象名来调用,结果发现不行。这个我可以理解,但第一种我就不能理解了。为什么protectedprivate的成员属性和方法在使用了__set__get这一对函数后在对象调用的时候就可以不用对象名呢?

不好意思我把题主问题里的代码修改了,我觉得题主的代码出现了一些思维问题,是因为题主完全对 PHP 面向对象了解太少。现在不得不使出八字箴言—— RTFMSTFW

题主的关键问题在于要理解 PHP 面向对象中 publicprotectedprivate 对于权限的影响,还有 PHP 魔术方法的原理。

给几个链接仅供题主参考:

http://www.php.net/manual/zh/language.oop5.overloading.php

http://www.php.net/manual/zh/language.oop5.magic.php

http://www.php.net/manual/zh/language.oop5.visibility.php

你是不是很奇怪为什么没创建的对象也还能对属性赋值?是不是?
如果真的是的话,我真觉得骚年你图样图森破啊!

你这里的$uu$g都不是person类实例化的对象好么,不信你执行var_dump()把三个都输出出来看看:

<code>var_dump($pef1);
var_dump($uu);
var_dump($g);
</code>
Copy after login

你会发现$pef1n1/n2/n3三个成员变量都有,而$uu$g中只有你赋值的属性。所以可以说明根本不是你所谓的person类实例化的对象好么!只是你恰好的用了n1属性而已。这段代码里头的$uu->n1$g->n1的作用基本等同于一个变量了,没有太大的讨论意义。

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