Is it possible to access outer local variable in PHP?(是否可以在 PHP 中访问外部局部变量?)
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.
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 :
而且,运行脚本,你会得到:
And, running the script, you'd get :
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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