Python Bruteforcing zip文件无法分配给函数调用

Posted 2023-03-06

【中文标题】Python Bruteforcing zip文件无法分配给函数调用

【英文标题】：Python Bruteforcing zip file cannot assign to function call

【发布时间】：2020-08-17 13:23:58

【问题描述】：

我正在学习如何使用 python BruteForcing 访问 zip 文件。但是当我在第 11 行的 zipF 中这样做时，我遇到了一个问题 这是一个例外：不能分配给函数调用。

import zipfile

zipF = zipfile.ZipFile
zipName = input("File path : ")
passwordFile = open("14MillionPass.txt","r")
for passw in passwordFile.readlines():
    ps = str(int(passw))
    ps = ps.encode()

try:
    with zipF.ZipFile(zipName) as myzip(): #the error is here
        myzip.extractAll(pwd = ps)
    print("Password found \n -> 0 is 1 password".format(ps,zipName))
    break
except:
    print("password not found")

提前致谢

【参考方案1】：

您不能在try-catch 语句中使用break。此外，您尝试将函数分配给文件处理程序。你可以用exit(0)代替break

try:
    with zipfile.ZipFile(zipName) as myzip:
        myzip.extractAll(pwd = ps)
    print("Password found \n -> 0 is 1 password".format(ps,zipName))
    exit(0) # successful exit
except:
    print("password not found")

而且你在你的程序中破坏了缩进，也许这是你想要的

import zipfile

zipName = input("File path : ")
passwordFile = open("14MillionPass.txt","r")
for passw in passwordFile.readlines():
    ps = str(int(passw))
    ps = ps.encode()
    try:
        with zipfile.ZipFile(zipName) as myzip:
            myzip.extractAll(pwd = ps)
        print("Password found \n -> 0 is 1 password".format(ps,zipName))
        break
    except:
        print("password not found")