Python Bruteforcing zip 文件无法分配给函数调用

Question

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 中执行此操作时遇到问题那是例外：无法分配给函数调用。

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

Answer 1

您不能在 try-catch 语句中使用 break。此外，您尝试为文件处理程序分配一个函数。您可以使用 exit(0) 而不是 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")

而且你的程序中的缩进被破坏了，也许这就是你想要的

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")

Python Bruteforcing zip 文件无法分配给函数调用

Python Bruteforcing zip file cannot assign to function call

python

zip

brute-force