Python 凯撒密码只对一个字母而不是整个字符串进行加密 所以我在做一个简单的凯撒密码练习，我不能让它破解整个字符串，只是单独的字母。< /P>

symbol\u add

是有问题的函数

代码如下：

import re

alphabet = "abcdefghijklmnopqrstuvwxyz"

def cleanIt(clean):
    global alphabet
    s = re.sub('[^a-z]+', '?', str(clean))
    return s

def symbol_add(symbol, key):
    encryptedMsg = ""
    for x in symbol:
        position = alphabet.find(x)
        newPosition = (position + key) % 26
        newLetter = alphabet[nyPosisjon]

    encryptedMsg += nyBokstav
    return encryptedMsg

def cipher(data,key):
    text = ""
    if data in alphabet:
        text += symbol_add(symbol=data,key=key)
        return text

def main():
    try:
        msg = (input("Write the message you would like to encrypt\n"))
        key = int(input("which key would you like to use?\n"))
        cleanIt(clean=msg)
        print(cipher(data=msg, key=key))
    except ValueError:
        print("Write a number!")

main()

我确信解决方案非常简单，仍在学习中

我们将非常感谢您对如何解决此问题的任何帮助

import re

alphabet = "abcdefghijklmnopqrstuvwxyz"

def cleanIt(clean):
    global alphabet
    s = re.sub('[^a-z]+', '?', str(clean))
    return s

def symbol_add(symbol, key):
    position = alphabet.find(symbol)
    newPosition = (position + key) % 26
    newLetter = alphabet[newPosition]
    return newLetter

def cipher(data,key):
    text = ""
    for letter in data:
        if letter in alphabet:
            text += symbol_add(symbol=letter,key=key)
    return text

def main():
    try:
        msg = (input("Write the message you would like to encrypt\n"))
        key = int(input("which key would you like to use?\n"))
        # Note: you need to assign msg to be equal to cleanIt(clean=msg). 
        #       Just calling cleanIt(clean=msg) won't work, as strings 
        #       are immutable in Python
        msg = cleanIt(clean=msg)
        print(cipher(data=msg, key=key))
    except ValueError:
        print("Write a number!")

main()

主要变化是：

• 循环的

  已从
  symbol\u add
  移动到
  cipher
  ，以便为每个字符调用
  symbol\u add

• 在

  main（）

  ：

  cleanIt（clean=msg）

  ->

  msg=cleanIt（clean=msg）

  ；这是因为字符串在Python中是不可变的，这意味着您需要重新分配变量

  msg

  ，以基本上指向新字符串

此代码的输出：

Write the message you would like to encrypt
test
which key would you like to use?
1
uftu

---

此外，尽量坚持单一的命名约定；您有一个函数跟在

camelCase

（

cleanIt

）后面，还有一个函数跟在

snake\u case

（

symbol\u add

）后面。尝试以相同的方式命名所有函数。（Python中的惯例是对函数使用

snake\u case

）

如果在输入密码方法时以查找方式填充字典，则可以大大简化密码方法。它包含基于提供的

键的映射，并将输入字符映射到密码字符

在dict中查找要比在字符串中查找
.index（）
快得多

使用
dict.get（key[，default]）
可以为未知项提供
“？”
，因此不需要
重新导入
，也不需要预处理

阅读有关dict.get（）的内容

：

基于小写映射，向chiffre添加大写映射也很简单：

alphabet = "abcdefghijklmnopqrstuvwxyz"

def cipher(data, key):
    # in case you change alphabet
    la = len(alphabet)

    # get the default lookup 
    chiffre = { c:alphabet[(i+key)%la] for i,c in enumerate(alphabet) }

    # create lookup for upper cases as well
    chiffre.update( { c.upper():n.upper() for c,n in chiffre.items() } )

    # supply ? for all unknowns, use the knowns where possible and return as string
    return ''.join( (chiffre.get(c,"?") for c in data) )

def main():
    try:
        msg = (input("Write the message you would like to encrypt\n"))
        key = int(input("which key would you like to use?\n"))

        print(cipher(data=msg, key=key))
    except ValueError:
        print("Write a number!")

main()

输出：

Write the message you would like to encrypt
Hello World
which key would you like to use?
1
Ifmmp?Xpsme

---

您的各种功能似乎对它们的操作感到困惑。您需要使每个函数一致地处理整个消息或单个字符谢谢！我知道传统的命名法，我是从另一种语言翻译过来的，目的是让它更容易理解和理解：）