凯撒密码加密/解密工具

Question

目前，我正在学习Python，我制作了这个脚本来锻炼我的技能。因为这是我的第一批程序之一，我希望得到一些反馈和提示，比如更多的组织、更少的代码行或更多的优化，任何我可以改进的东西。

print "Do you want to encrypt or decrypt the message (e/d)"
action = raw_input("> ")
while action != "e" and action != "d":
    action = raw_input( "(e/d): ")

abc_l_e = "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz"
abc_u_e = "ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ"
abc_l_d = "zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba"
abc_u_d = "ZYXWVUTSRQPONMLKJIHGFEDCBAZYXWVUTSRQPONMLKJIHGFEDCBA"
chars = "!\"#$%&'()*+-'-./:;<=>?@[]^_{}|~1234567890"

if action == "e":
    print "\nEnter the message to encrypt"
    message = raw_input("> ")

    print "\nEnter the key"
    try:
        key = input("> ")
        while not (key >= 1 and key <= 26):
            print "Number not in range (1-26). Try again"
            key = input("> ")
    except:
        print "That's not a correct integer. Try again."
        exit()

    e_message = ""
    for char in message:
        if char in abc_l_e:
            e_message += abc_l_e[(abc_l_e.index(char) + key)]
        if char in abc_u_e:
            e_message += abc_u_e[(abc_u_e.index(char) + key)]
        elif char == " ":
            e_message += " "
        elif char in chars:
            e_message += char
    print "Your encrypted message is:\n", e_message

elif action == "d":
    print "\nEnter the message to decrypt"
    message = raw_input("> ")
    print "\nEnter the key (enter \"n\" if you don't know)"
    try:
        key = raw_input("> ")
        if key != "n":
            while not (int(key) >= 1 and int(key) <= 26):
                print "Number not in range (1-26). Try again"
                key = input("> ") 
    except:
        print "That's not a correct integer nor \"n\". Try again."
        exit()

    d_message = ""
    if key != "n":
        key = int(key)  
        for char in message:
            if char in abc_l_d:         
                d_message += abc_l_d[(abc_l_d.index(char) + key)]
            elif char in abc_u_d:
                d_message += abc_u_d[(abc_u_d.index(char) + key)]
            elif char == " ":
                d_message += char
            elif char in chars:
                d_message += char
        print "Your decrypted message is:\n", d_message

    if key == "n":
        results = []
        for turn in range(1,26):
            d_message = ""
            for char in message:
                if char in abc_l_d:
                    d_message += abc_l_d[(abc_l_d.index(char) + turn)]
                elif char in abc_u_d:   
                    d_message += abc_u_d[(abc_u_d.index(char) + turn)]
                elif char == " ":
                    d_message += char
                elif char in chars:
                    d_message += char
            results.append(d_message)

        print "\nYour decrypted message is between one of the following:"
        for result in results:
            print "%d." % (results.index(result) + 1), result

Answer 1

• 与其提示输入，几乎所有shell脚本(包括此脚本)都将通过接受静态输入(动作和键)作为参数，然后将标准输入读取到EOF以生成输出，从而得到改进。为此，您将需要一个main方法，它只是编排参数解析，并将结果值传递给执行实际替换的函数或方法。这使得以一种常见的方式将脚本与其他命令组合起来非常容易。argparse非常适合这一点。
• 加密和解密只是将字符左转右移的问题。您可以使用这些信息将代码减半。
• 蛮力法应该通过运行一个函数26次来实现。
• 一般来说，代码应该被分解成函数，直到任何地方都没有明显的重复。这将极大地简化除了输入处理之外的所有内容，这意味着脚本中有趣的部分最终应该是三行。
• 您应该使用一次只处理一行输入，这样即使是巨大的输入也会导致小内存的使用。