是否可以在 PHP 中访问外部局部变量?

2022-01-04 00:00:00 scope php

是否可以在 PHP 子函数中访问外部局部变量?

Is it possible to access outer local varialbe in a PHP sub-function?

在下面的代码中，我想访问内部函数栏中的变量 $l.在栏中将 $l 声明为 global $l 不起作用.

In below code, I want to access variable $l in inner function bar. Declaring $l as global $l in bar doesn't work.

function foo()
{
    $l = "xyz";

    function bar()
    {
        echo $l;
    }
    bar();
}
foo();

推荐答案

你或许可以使用闭包来做到这一点......

You could probably use a Closure, to do just that...


花了一些时间来记住语法，但它看起来像这样:


Edit : took some time to remember the syntax, but here's what it would look like :

function foo()
{
    $l = "xyz";
    $bar = function () use ($l)
    {
        var_dump($l);
    };
    $bar();
}
foo();

而且，运行脚本，你会得到:

And, running the script, you'd get :

$ php temp.php
string(3) "xyz"


一些注意事项:


A couple of note :

  • 你必须在函数声明之后放一个;！
  • 你可以通过引用use变量，在它的名字前加上一个&:use (& $l)
  • You must put a ; after the function's declaration !
  • You could use the variable by reference, with a & before it's name : use (& $l)

更多信息，作为参考，你可以看看手册中的这个页面:匿名函数

For more informations, as a reference, you can take a look at this page in the manual : Anonymous functions

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