Python ZIP压缩文件破解

# -*- coding:utf-8 -*-

import zipfile
import os
import string
import random


class crackzip():
    def __init__(self, low=4, up=6):
        self.lownum = low
        self.upnum = up
        # 构造密码的字符串
        self.model = string.ascii_lowercase + '0123456789'
        # ZIP文件位置
        self.in_path = "C:\Users\10245\Desktop\test3.zip"
        # 解压位置
        self.out_path = "C:\Users\10245\Desktop\re\"

    def __iter__(self):
        return self

    def __next__(self):
        s = ''
        for i in range(random.randrange(self.lownum, self.upnum)):
            s += random.choice(self.model)
        return s

    def dict_zip(self):
        # 字典文档位置
        dic = "C:\Users\10245\Desktop\dict.txt"
        m = zipfile.ZipFile(self.in_path)
        names = [m.filelist[x].filename for x in range(len(m.filelist))]
        corr_name = []
        for name in names:
            corr_name.append(name.encode('cp437').decode('gbk'))
        with open(dic) as f:
            passwords = f.readlines()
        for password in passwords:
            try:
                password = password.strip()
                m.extractall(path=self.out_path, pwd=password.encode('utf-8'))
                for n in range(len(names)):
                    os.rename(self.out_path + names[n], self.out_path + corr_name[n])
                print ('[+][Success]文件：{0}  密码：{1}'.format(corr_name, password))
                exit(0)
            except Exception as e:
                print ('[-][Error]文件：{0}  密码：{1}'.format(corr_name, password))

    def str_zip(self):
        m = zipfile.ZipFile(self.in_path)
        # zip中的文件
        names = [m.filelist[x].filename for x in range(len(m.filelist))]
        corr_name = []
        sec_num1, sec_num2 = map(int, input('请输入密码位数范围：').split())
        for name in names:
            corr_name.append(name.encode('cp437').decode('gbk'))
        for password in crackzip(sec_num1, sec_num2):
            try:
                password = password.strip()
                m.extractall(path=self.out_path, pwd=password.encode('utf-8'))
                for n in range(len(names)):
                    os.rename(self.out_path + names[n], self.out_path + corr_name[n])
                print ('[+][Success]文件：{0}  密码：{1}'.format(corr_name, password))
                exit(0)
            except Exception as e:
                print ('[-][Error]文件：{0}  密码：{1}'.format(corr_name, password))
        m.close()


if __name__ == '__main__':
    print ('''
    (1) 字典破解
    (2) 常用构造破解
    ''')
    m = crackzip()
    ch = int(input())
    if ch == 1:
        m.dict_zip()
    elif ch == 2:
        m.str_zip()

字典破解，就是尝试常用密码文本中的密码。

暴力破解，使用字符串构造密码，Python性能不咋样，所以没法直接构造全部密码，再一一尝试，用了迭代器，每次随机取值。