请注意,本文编写于 1519 天前,最后修改于 1516 天前,其中某些信息可能已经过时。
做题时碰到过这类的截断验证,转载过来方便日后查看。原文链接:https://blog.csdn.net/lacoucou/article/details/72355346
ctf中经常用MD5的截断比较做验证,如:substr(md5($str), 0, 6) === “3322cf”,通过这种方式限制脚本的自动化攻击。
通常可以写脚本去爆破MD5, 但是花费时间和比较字符串的长度有关,并且花费时间通常比较长,这不利于脚本自动攻击,下面给出爆破脚本和使用方式。
submd5.py
# -*- coding: utf-8 -*-
import multiprocessing
import hashlib
import random
import string
import sys
CHARS = string.letters + string.digits
def cmp_md5(substr, stop_event, str_len, start=0, size=20):
global CHARS
while not stop_event.is_set():
rnds = ''.join(random.choice(CHARS) for _ in range(size))
md5 = hashlib.md5(rnds)
if md5.hexdigest()[start: start+str_len] == substr:
print rnds
stop_event.set()
if __name__ == '__main__':
substr = sys.argv[1].strip()
start_pos = int(sys.argv[2]) if len(sys.argv) > 1 else 0
str_len = len(substr)
cpus = multiprocessing.cpu_count()
stop_event = multiprocessing.Event()
processes = [multiprocessing.Process(target=cmp_md5, args=(substr,
stop_event, str_len, start_pos))
for i in range(cpus)]
for p in processes:
p.start()
for p in processes:
p.join()
用法:
#python submd5.py substr startpos
$ python submd5.py "3d4f4"
SponhjOhIZ30IaM1fweb
$ python submd5.py "3df4" 2
G8tr6VhonA1z3xJdaGBu
想要在较短时间内获得可用的md5,可以使用彩虹表类似的方式去实现,通过空间去换时间。
生成md5文件,并排序:
gen_md5.py
# -*- coding: utf-8 -*-
import itertools
import hashlib
import string
CHARS = string.letters + string.digits
str_len = 8
for s in itertools.product(CHARS, repeat=str_len):
s = ''.join(s)
print "{0} {1}".format(hashlib.md5(s).hexdigest(), s)
命令行排序:
python gen_md5.py | sort -o md5_sorted.txt
md5 文件搜索(二分查找):
match.py
# -*- coding: utf-8 -*-
import itertools
import hashlib
import string
import os
def match(s):
md5_file = "md5_sorted.txt"
byte_size = os.path.getsize(md5_file)
with open(md5_file, 'rb') as f:
line_len = len(f.readline())
print line_len
with open(md5_file, "rb") as f:
L = 0
R = byte_size / line_len - 1
while R - L > 0:
C = L + (R - L) / 2
offset = C * line_len
f.seek(offset)
ln = f.read(line_len).strip()
#print ln
head = ln[:len(s)]
if s == head:
return ln.split(" ")[1]
if s < head:
R = C
continue
L = C
return
# print match('fef')