python暴力解压rar加密文件过程详解
作者:_qiufeng 时间:2023-11-20 06:28:38
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()
来源:https://blog.csdn.net/weixin_43873887/article/details/87862831
标签:python,暴力,解压,rar,加密,文件
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