使用 Python 破解压缩文件的密码的思路详解

import itertools
def allkeyword(dic,num):
 allkey1 = itertools.product(dic,repeat=num)
 allkey2 = (''.join(i) for i in allkey1)
 return allkey2

dictionaries = ['1', '2']
print(list(allkeyword(dictionaries,4)))
# ['1111', '1112', '1121', '1122', '1211', '1212', '1221', '1222', '2111', '2112', '2121', '2122', '2211', '2212', '2221', '2222']

import zipfile
try:
 ZIPFILE = zipfile.ZipFile(r'D:\123\1.zip') # 注意路径
 ZIPFILE.extractall(path=r'D:\123',pwd=b'1234') # 解压到哪个路径
 print("解压成功")

except:
 print("解压失败")

import zipfile
import itertools
# 破解一个4位数密码数字和字母为23ab大概5-10分钟，仅供参考。
dictionaries = ['1', '2', '3', '4','5','6','7','8','9','0',
    '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']   #组成破解字典的关键字符（可以按照自己需求添加）
end_for = True  # 用于破解成功后，停止循环的变量
# 设置密码的长度1到16位密码
for x in range (1,17):
 if end_for:
  def allkeyword():
   allkey1 = itertools.product(dictionaries,repeat=x)
   allkey2 = (''.join(i) for i in allkey1)
   return allkey2

  def trypassword (password):
   try:
    ZIPFILE = zipfile.ZipFile(r'D:\zip\abc.zip') # 需要解压带有密码的本地abc.zip
    ZIPFILE.extractall(path=r'D:\zip',pwd=password.encode('utf-8'))  # 解压到哪个路径下
    print(f"解压成功,正确密码为：{password}")  # 解压成功，并打印出正确密码
    global end_for  # 声明为全局变量，没有声明，重新赋值无效
    end_for = False  # 解压成功，停止循环
    return True
   except:
    print(f"解压失败,尝试密码为：{password}") 
    return False

  #用trypassword函数返回的True或者Flase来判定程序是否终止。
  for pwd in allkeyword() : 
   if trypassword(pwd):
    break