PHP 类:全局变量作为类中的属性

2021-12-23 00:00:00 class variables methods php

我在班级之外有一个全局变量 = $MyNumber;

I have a global variable outside my class = $MyNumber;


我如何将其声明为 myClass 中的属性?
对于我班上的每个方法,我都是这样做的:


How do I declare this as a property in myClass?
For every method in my class, this is what I do:

class myClass() {

    private function foo() {
         $privateNumber = $GLOBALS['MyNumber'];
    }

}



我想要这个



I want this

class myClass() {

    //What goes here?
    var $classNumber = ???//the global $MyNumber;

    private function foo() {
         $privateNumber = $this->classNumber;
    }

}



我想基于全局 $MyNumber 创建一个变量,但是
在方法中使用之前修改

I want to create a variable based on the global $MyNumber but
modified before using it in the methods

类似: var $classNumber = global $MyNumber + 100;

something like: var $classNumber = global $MyNumber + 100;

推荐答案

您可能不真的想这样做,因为调试将是一场噩梦,但它似乎成为可能.关键是您在构造函数中通过 reference 分配的部分.

You probably don't really want to be doing this, as it's going to be a nightmare to debug, but it seems to be possible. The key is the part where you assign by reference in the constructor.

$GLOBALS = array(
    'MyNumber' => 1
);

class Foo {
    protected $glob;

    public function __construct() {
        global $GLOBALS;
        $this->glob =& $GLOBALS;
    }

    public function getGlob() {
        return $this->glob['MyNumber'];
    }
}

$f = new Foo;

echo $f->getGlob() . "
";
$GLOBALS['MyNumber'] = 2;
echo $f->getGlob() . "
";

输出将是

1
2

这表明它是通过引用而不是值来分配的.

which indicates that it's being assigned by reference, not value.

正如我所说,调试将是一场噩梦,所以你真的不应该这样做.通读维基百科关于封装的文章;基本上,理想情况下,您的对象应该管理自己的数据以及修改该数据的方法;恕我直言,即使是公共财产也是个坏主意.

As I said, it will be a nightmare to debug, so you really shouldn't do this. Have a read through the wikipedia article on encapsulation; basically, your object should ideally manage its own data and the methods in which that data is modified; even public properties are generally, IMHO, a bad idea.

相关文章