无法杀死我的python代码.怎么了？

2019年8月3日 199次阅读

好的,所以我在
python中编写了一个非常简单的密码破解程序,强制使用字母数字字符强制密码.目前,此代码仅支持1个字符的密码和密码文件,其中包含md5哈希密码.它最终将包含指定您自己的字符限制的选项(破解程序在失败之前尝试的字符数).现在,当我希望它死时,我无法杀死这段代码.我已经包含了一个尝试,除了snippit,但它不起作用.我做错了什么？

代码：http://pastebin.com/MkJGmmDU

import linecache, hashlib

alphaNumeric = ["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","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",1,2,3,4,5,6,7,8,9,0]

class main:
    def checker():
            try:
                    while 1:
                            if hashlib.md5(alphaNumeric[num1]) == passwordHash:
                                    print "Success! Your password is: " + str(alphaNumeric[num1])
                                    break
            except KeyboardInterrupt:
                    print "Keyboard Interrupt."

    global num1, passwordHash, fileToCrack, numOfChars
    print "What file do you want to crack?"
    fileToCrack = raw_input("> ")
    print "How many characters do you want to try?"
    numOfChars = raw_input("> ")
    print "Scanning file..."
    passwordHash = linecache.getline(fileToCrack, 1)[0:32]
    num1 = 0

    checker()

main

最佳答案 允许KeyboardInterrupt结束程序的方法是什么都不做.他们的工作依赖于在一个除外块中捕捉它们的任何东西;当一个异常从程序(或线程)中一路消失时,它就会终止.

你所做的是捕获KeyboardInterrupts并通过打印消息然后继续来处理它们.

至于程序卡住的原因,没有任何东西可以导致num1改变,因此md5计算每次都是相同的计算.如果你想迭代alphaNumeric中的符号,那么这样做：对于alphaNumeric中的符号：#用’symbol’做一些事情.

当然,这仍然只考虑每个可能的单字符密码.你将不得不努力尝试…… 🙂

我觉得你也对课程的使用感到困惑. Python不要求你将所有内容都包装在一个类中.程序结尾处的主要内容没有任何用处;您的代码运行是因为在编译器试图找出主类是什么时会对其进行评估.这是滥用语法.您要做的是将此代码放在main函数中,并调用该函数(与您当前调用checker的方式相同).

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