是否可以在PHP中访问外部局部变量? [英] Is it possible to access outer local variable in PHP?

问题描述

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，我想访问内部功能栏中的变量$l.在bar中将$l声明为global $l无效.

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

您可能可以使用Closure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，但这是这样的:

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();

然后，运行脚本，您将得到:

And, running the script, you'd get :

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

一些注意事项:

A couple of note :

• 您必须在函数的声明后加上;！
• 您可以通过引用use将该变量命名为use (& $l)
，并在其前加上&.

• You must put a ; after the function's declaration !
• You could use the variable by reference, with a & before it's name : use (& $l)

有关更多信息，请参考手册中的以下页面: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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