从包含文件返回 [英] Return from include file

问题描述

在PHP中，如何从包含的脚本返回到包含该脚本的脚本?

in PHP, how would one return from an included script back to the script where it had been included from?

IE:

1-主脚本 2-申请 3-包含

1 - main script 2 - application 3 - included

基本上，我想从3变回2，return()不起作用.

Basically, I want to get back from 3 to 2, return() doesn't work.

代码2-应用

$page = "User Manager";
if($permission["13"] !=='1'){
    include("/home/radonsys/public_html/global/error/permerror.php");
    return();
}

推荐答案

includeme.php:

includeme.php:

$x = 5;
return $x;

main.php:

$myX = require 'includeme.php'; 

也给出相同的结果

这是PHP鲜为人知的功能之一，但是对于设置非常简单的配置文件来说可能会很好.

This is one of those little-known features of PHP, but it can be kind of nice for setting up really simple config files.

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