Question

查看此代码，如果我希望此代码从wordlist中随机获取字词，直到获得名为flag的内容，我该怎么办？

from Crypto.Cipher import AES
import base64
import os

BLOCK_SIZE = 32

PADDING = '{'

# Encrypted text to decrypt
encrypted = "Z5p+ZK9f8m9z+wVHw2SsvS0qT0DtqiTY1+yStCzXvP4="

DecodeAES = lambda c, e:   c.decrypt(base64.b64decode(e)).rstrip(PADDING)

secrets = #here you need to open words.txt

for secret in secrets:
if (secret[-1:] == "\n"):
    print("Error, new line character at the end of the string. This will not match!")
elif (len(secret) >= 32):
    print ("Error, string too long. Must be less than 32 characters.")

else:
    # create a cipher object using the secret
    cipher = AES.new(secret + (BLOCK_SIZE - len(secret) % BLOCK_SIZE) * PADDING)

    # decode the encoded string
    decoded = DecodeAES(cipher, encrypted)

    if (decoded.startswith('FLAG:')):
        print ("\n")
        print ("Success: "+secret+"\n")
        print (decoded+"\n")
    else:
        print ('Wrong password')

我尝试使用模块导入它并仍然遇到同样的问题

Answer 1

当您说secrets时，您将获得所有可能性，因为文本文件中的每一行都将以换行符结尾。您可以将其取出并将[line.strip() for line in open("secrets.txt")]定义为import javax.swing.JOptionPane; import java.util.Date; import java.text.*; public class JOptionQuestionTen { public static void main(String[] args) { String input = JOptionPane.showInputDialog("Insert Date in Form MM/dd/yy"); DateFormat inputFormat = new SimpleDateFormat("MM/dd/yy"); DateFormat outputFormat = new SimpleDateFormat("dd.MM.yy"); Date date = inputFormat.parse(input); String formattedDate = outputFormat.format(date); JOptionPane.showMessageDialog(null, formattedDate); } }

Answer 2

我假设您知道如何打开和阅读文件;在许多在线资料中找到这些都是微不足道的。

请注意，您的代码会按顺序遍历单词列表。要随机获取条目，请尝试以下操作：

import random

word_list = [
    "apple",
    "baker",
    "cat",
    "FLAG:",
    "gorilla"
]

word = random.choice(word_list)
while not word.startswith("FLAG:"):
    print word
    word = random.choice(word_list)

一次长跑的输出：

cat
apple
apple
gorilla
cat
cat
baker
cat
FLAG: stop here 
   DONE

在linux

2 个答案: