python暴力解压rar加密文件过程详解

Hasana ·
更新时间:2024-11-14
· 714 次阅读

第一次使用csdn写文章,写得不好还请见谅。(运行环境:python3.6)

下了一个带密码的压缩包文件,作为一个刚学python的新手,想着能不能用python暴力破解它,于是在网上搜了很多资料,看着似乎并不是很麻烦,也想试着自己写一个可以暴力破解的程序,在写的过程中却遇到了各种各样的问题,希望大手们能带带我。遇到的问题如下:

zipfile和zipfile2似乎都不支持AES解密(https://bugs.python.org/issue9170) 在用rarfile暴力破解时即使密码错误也不抛出异常,因此无法用try,except捕获密码

本来是想写一个可以同时暴力破解zip和rar的程序,在试了半天解密zip却一直提示密码错误之后放弃了zip,想着能不能写一个暴力破解rar的程序。

首先是生成字典:要用到itertools模块

import itertools as its import string def createDict(path,repeats,words): dict = its.product(words,repeat=repeats) '''这里的words是要迭代的字符串,repeats是生成的密码长度,生成的dict是一个返回元组的迭代器''' f = open(path,'a') for cipher in dict: f.write(''.join(cipher) + '\n') f.close() def main(): numbers = string.digits #包含0-9的字符串 path = '输入你的字典路径' length = 你要迭代的密码长度 for i in range(1,length): createDict(path,i,numbers) if __name__=="__main__": main()

到这里我们的字典已经生成完毕了,接下来开始暴力破解rar

from threading import Thread from unrar import rarfile import os '''首先我们要读取字典,字典可能太大因此我们采用迭代器''' def get_pwd(dict_path): with open(dict_path,'r') as f: for pwd in f: yield pwd.strip() def decode_rar(fp,pwd,extract_path): try: fp.extractall(extract_path,pwd=pwd) except: pass else: print('the pwd is>',pwd) ''' 事实上我在尝试时似乎从来没有到达过else,这样可能是得不到解压密码的。我的 一种得到密码的想法如下,但是运行效率可能会降低 def decode_rar(fp,pwd,check_file,extract_path): fp.extractall(extract_path,pwd=pwd) if os.path.exists(check_file): print('The pwd is:',pwd) exit(0) 其中check_file可以设置为fp.namelist()[0] 并且该方法不能使用多线程,因此速度会降低很多 ''' def main(): extract_path = '你要解压的路径' dict_path = '你的字典路径' filename = '你的rar路径' fp = rarfile.RarFile(filename) pwds = get_pwd(dict) '''使用多线程可提高速度''' for pwd in pwds: t = Thread(target=rar_file,args=(fp,pwd,extract_path)) t.start()

以上是写程序的思路和遇到的各种坑,代码是手敲的,可能有一些错误,希望能得到谅解和帮助。

下面是一个图形界面的rar解密源代码:(图形只是想练习,运行较慢,建议直接运行上面的函数)

import tkinter as tk import os from tkinter import messagebox from unrar import rarfile from threading import Thread def getPwd(dict): with open(dict,'r') as f: for pwd in f: yield pwd.strip() def slowerDecode(fp,pwd,check_file,extract_path): fp.extractall(extract_path,pwd=pwd) if os.path.exists(check_file): messagebox.showinfo(message="密码:"+pwd) messagebox.showinfo(message="程序结束") messagebox.showinfo(message="密码:"+pwd) exit(0) def quickDecode(fp,pwd,extract_path): fp.extractall(extract_path,pwd=pwd) def check(obs): flag = 1 for ob in obs: if not ob.checkExist(): flag = 0 ob.showError() if(not flag): return 0 else: for ob in obs: if not ob.check(): flag = 0 ob.showError() if (not flag): return 0 else: for ob in obs: ob.right() return 1 def main(obs): extract_path = obs[0].path_input.get() rar_path = obs[1].path_input.get() txt_path = obs[2].path_input.get() pwds = getPwd(txt_path) global var1 global var2 if(check(obs)): if(var1.get() == 0 and var2.get() == 0): messagebox.showerror(message="选择一个选项!!!") elif(var1.get() == 0 and var2.get() == 1): fp = rarfile.RarFile(rar_path) check_file = fp.namelist()[0] for pwd in pwds: slowerDecode(fp,pwd,check_file,extract_path) elif(var1.get() == 1 and var2.get() == 0): fp = rarfile.RarFile(rar_path) for pwd in pwds: t = Thread(target=quickDecode,args=(fp,pwd,extract_path)) t.start() exit(0) else: messagebox.showerror(message="只选择一个!!!") class FolderPath: def __init__(self,y=0,error_message="Not exists!",path_input="",text=''): self.y = y self.error_message = error_message self.path_input = path_input self.text = text def createLabel(self): label = tk.Label(window,bg="white",font=("楷体",13),width=20,text=self.text) cv.create_window(100,self.y,window=label) def createEntry(self): entry = tk.Entry(window,fg="blue",width="40",bg="#ffe1ff",textvariable=self.path_input) cv.create_window(330,self.y,window=entry) def show(self): self.createLabel() self.createEntry() def showError(self,color="red"): label = tk.Label(window,bg="white",fg=color,font=("楷体",13),width="10",text=self.error_message) cv.create_window(530,self.y,window=label) def checkExist(self): self.error_message = 'Not exists!' if not os.path.exists(self.path_input.get()): return 0 return 1 def check(self): if not os.path.isdir(self.path_input.get()): self.error_message = 'Not a dir!' return 0 else: return 1 def right(self): self.error_message = "right path!" self.showError('#00FFFF') class FilePath(FolderPath): def check(self): if (self.path_input.get().split('.')[-1] == self.suffix): return 1 else: self.error_message = "Not "+self.suffix + '!' return 0 window = tk.Tk() window.title('made by qiufeng') window.geometry('600x300') cv = tk.Canvas(window,width=600,height=300,bg='white') cv.pack() folderpath = FolderPath(y=140,path_input=tk.StringVar(),text="请输入解压路径") folderpath.show() rarpath = FilePath(y=60,path_input=tk.StringVar(),text="请输入rar路径") rarpath.suffix = 'rar' rarpath.show() txtpath = FilePath(y=100,path_input=tk.StringVar(),text="请输入字典路径") txtpath.suffix = 'txt' txtpath.show() obs = [folderpath,rarpath,txtpath] #多选框 var1 = tk.IntVar() var2 = tk.IntVar() ck1 = tk.Checkbutton(window,text="直接破解(无法获得密码)",variable=var1) cv.create_window(150,200,window=ck1) ck2 = tk.Checkbutton(window,text="慢速(可获得密码)",variable=var2) cv.create_window(132,230,window=ck2) button = tk.Button(window,text="确认",command=lambda: main(obs)) cv.create_window(90,260,window=button) window.mainloop() 您可能感兴趣的文章:Python编译成.so文件进行加密后调用的实现Python中使用pypdf2合并、分割、加密pdf文件的代码详解python破解zip加密文件的方法Python基于hashlib模块的文件MD5一致性加密验证示例python文件的md5加密方法RC4文件加密的python实现方法python实现文件快照加密保护的方法基于python实现文件加密功能



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