Discuz/PHPBB/phpcmsv9等破解脚本

对于md5(md5(pass).slat)这样的加密,破解的难度大大增加,前面发了几篇后台的绕过,都需要session有效的情况下才行。

网上找了下 有人写了这类的脚本,但是是php版本的,速度不行,自己写了一个多线程的破解脚本。

#!/usr/bin/env python
# -*- coding: cp936 -*-
#code by 花开、若相惜
#Pax.Mac Team
import hashlib
import Queue
from threading import Thread
import sys
import time


class End():
    def __init__(self):
        self.end = False
         
    def Finish(self):
        self.end = True
     
    def GetEnd(self):
        return self.end 
 
class Connection(Thread):
    def __init__(self,queue,TheEnd):
        Thread.__init__(self)
        self.queue = queue
        self.TheEnd = TheEnd
        
    def run(self):
        while (not self.TheEnd.GetEnd()) and (not self.queue.empty()):
            pwd = self.queue.get()
            if exploit(pwd)==1:
                print pwd
                self.TheEnd.Finish()

def exploit(password):
    encrypt="paxmac"; 
    passmd5="232997126b4cbc244997d7c11136d469";
    hashmd5=hashlib.md5((hashlib.md5(password).hexdigest()+encrypt)).hexdigest()
    if hashmd5==passmd5:
        res=1
    else:
        res=0
    return res


def main():

    queue=Queue.Queue()
    TheEnd = End()
    pwd_count = len(open('password.txt','rU').readlines())
    print '你的密码条数：%d'%(pwd_count)
    pwds = [line.rstrip() for line in open("password.txt")]
    for pwd in pwds:
        queue.put(pwd)
    initsize = queue.qsize()
    tested = 0
    threads = 1    #修改线程处
    for i in range(0,int(threads)):
        Connection(queue,TheEnd).start()
    while (not TheEnd.GetEnd()) and (not queue.empty()):
        time.sleep(2)
        actsize = queue.qsize()
        speed = (initsize - tested - actsize) / 2
        tested = initsize - actsize    
        print 'use %i password | Remaining %i password | Speed %i/sec' %(tested,actsize,speed)
        
if __name__ == '__main__':
    main()

 

 

 

大概8.8W/S左右的破解速度,实在不行的情况下可以试试。

crackhash_1