说明:

编辑:我只需要方法菜单返回“即将到来”,因为-就目前而言,如果用户输入c,p或s,则它什么也不返回。我不明白为什么会这样。

def PrintDescription():
    print 'This program encrypts and descrypts messages using multiple \
encryption methods.\nInput files must be in the same directory as this program.\
\nOutput files will be created in this same directory.'

def StartMenu():
    print 'Do you wish to encrypt or decrypt?'
    print '<e>ncrypt'
    print '<d>ecrypt'
    print '<q>uit'

def MethodMenu():
  print 'Which method would you like to use?'
  print '<c>aesarian fixed offset'
  print '<p>seudo-random offset'
  print '<s>ubstitution cipher'
  a = raw_input("")
  while a not in ('c', 'p', 's'):
    if a:
      print "Error: You must type c, p, or s"
      a = raw_input("")
    if a == 'c' or a=='p' or a=='s':
      print 'Coming Soon'

def main():
    alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789,.?! \t\n\r"
    PrintDescription()
    a = None
    while a not in ('e', 'd', 'q'):
        if a:
            print "Error: You must type e, d, or q"
        else:
            StartMenu()
        a = raw_input("")
        if a == 'e' or a=='d':
          MethodMenu()
        if a == 'q':
          break

main()

最佳答案

在提出解决方案之前,这里有一些评论。


MethodMenu()函数当前不返回任何内容。我认为您是要退还用户的选择。
我看到了StartMenu()和MethodMenu()之间的模式:每个模式都显示一个选择列表,并反复获取用户的输入,直到用户输入正确的输入为止。但是,StartMenu()函数不管理用户的输入,而MethodMenu()确实在设计中==>不一致。
由于获取用户输入并对其进行验证的行为发生了两次,因此最好将代码块移到一个单独的函数中,您可以调用该函数,而不是多次编写同一代码块。
我注意到单字母变量a的用户。通常,我建议使用更具描述性的名称,例如user_choice,user_answer或user_input。


事不宜迟,我的解决方案是:

def PrintDescription():
    print 'This program encrypts and descrypts messages using multiple \
encryption methods.\nInput files must be in the same directory as this program.\
\nOutput files will be created in this same directory.'

def GetChoice(acceptable_answers):
    while True:
        user_choice = raw_input('')
        if user_choice in acceptable_answers:
            return user_choice
        else:
            print 'Please try:', ', '.join(acceptable_answers)

def StartMenu():
    print 'Do you wish to encrypt or decrypt?'
    print '<e>ncrypt'
    print '<d>ecrypt'
    print '<q>uit'
    user_choice = GetChoice('edq')
    return user_choice

def MethodMenu():
    print 'Which method would you like to use?'
    print '<c>aesarian fixed offset'
    print '<p>seudo-random offset'
    print '<s>ubstitution cipher'
    user_choice = GetChoice('cps')
    return user_choice

def main():
    alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789,.?! \t\n\r"
    PrintDescription()

    while True:
        user_choice = StartMenu()
        if user_choice in ('e', 'd'):
            user_choice = MethodMenu()
            # Do something based on the user_choice
        if user_choice == 'q':
            break

main()


更新资料

如果您必须了解MethodMenu()的问题,请按以下说明进行操作:用户第一次键入正确的选项(c,p或s):跳过了整个while循环,这意味着“即将推出”被打印。您可以修改解决方案,也可以使用hek2mgl。

10-08 15:11