在 PHP 函数中访问全局变量

问题描述

根据大多数编程语言的作用域规则，我可以访问在函数内部定义的变量，但为什么这段代码不起作用?

According to the most programming languages scope rules, I can access variables that are defined outside of functions inside them, but why doesn't this code work?

<?php
    $data = 'My data';

    function menugen() {
        echo "[" . $data . "]";
    }

    menugen();
?>

输出为[].

推荐答案

这是一个范围问题.简而言之，应该避免使用全局变量所以:

It's a matter of scope. In short, global variables should be avoided so:

您要么需要将其作为参数传递:

You either need to pass it as a parameter:

$data = 'My data';

function menugen($data)
{
    echo $data;
}

或者把它放在一个类中并访问它

Or have it in a class and access it

class MyClass
{
    private $data = "";

    function menugen()
    {
        echo this->data;
    }

}

另请参阅@MatteoTassinari 答案，因为您可以将其标记为全局以访问它，但通常不需要全局变量，因此重新考虑您的编码是明智的.

See @MatteoTassinari answer as well, as you can mark it as global to access it, but global variables are generally not required, so it would be wise to re-think your coding.

这篇关于在 PHP 函数中访问全局变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！