我试图完成我的程序我添加了一个菜单,允许用户选择一些选项,允许用户存储网站名称和密码在列表中。但是,当我将一些网站名称和密码添加到它们各自的保险库中时,出现了一个问题,当我试图在附加网站名称和密码后选择一个选项时,“1”是调用viewapp()函数以查看到目前为止存储的网站和密码的预期输入。问题是调用viewapp()函数需要两次以上的时间,它拒绝第一个预期输入,但奇怪地接受第二个输入。另外,当我选择第三个选项调用summary()时,整个打印的摘要将打印两次,这与只接受第二个预期输入的菜单模式类似。这个程序正在做我想做的,除了这个讨厌的错误,选择这四个选项会让它第二次请求输入,而它应该直接跳到那个函数。如有帮助,不胜感激。在
appvault = []
passvault = []
def logged():
print("----------------------------------------------------------------------\n")
print("Hello, welcome to the password vault console. ")
modea = input("""Below are the options you can choose from in the password vault console:
##########################################################################\n
1) Find the password for an existing webiste/app
2) Add a new website/app and a new password for it
3) Summary of the password vault
4) Exit
##########################################################################\n
> """).strip()
return modea
def viewapp():
if len(appvault) > 0:
for app in appvault:
print("Here is the website/app you have stored:")
print("- {}\n".format(app))
if len(passvault) > 0 :
for code in passvault:
print("Here is the password you have stored for the website/app: ")
print("- {}\n".format(code))
else:
print("You have no apps or passwords entered yet!")
def addapp():
while True:
validapp = True
while validapp:
new_app = input("Enter the new website/app name: ").strip().lower()
if len(new_app) > 20:
print("Please enter a new website/app name no more than 20 characters: ")
elif len(new_app) < 1:
print("Please enter a valid new website/app name: ")
else:
validapp = False
appvault.append(new_app)
validnewpass = True
while validnewpass:
new_pass = input("Enter a new password to be stored in the passsword vault: ")
if not new_pass.isalnum():
print("Your password for the website/app cannot be null, contain spaces or contain symbols \n")
elif len(new_pass) < 8:
print("Your new password must be at least 8 characters long: ")
elif len(new_pass) > 20:
print("Your new password cannot be over 20 characters long: ")
else:
validnewpass = False
passvault.append(new_pass)
validquit = True
while validquit:
quit = input("\nEnter 'end' to exit or any key to continue to add more website/app names and passwords for them: \n> ")
if quit in ["end", "End", "END"]:
logged()
else:
validquit = False
addapp()
return addapp
def summary():
if len(passvault) > 0:
for passw in passvault:
print("----------------------------------------------------------------------")
print("Here is a summary of the passwords stored in the password vault:\n")
print("The number of passwords stored:", len(passvault))
print("Passwords with the longest characters: ", max(new_pass for (new_pass) in passvault))
print("Passwords with the shortest charactrs: ", min(new_pass for (new_pass) in passvault))
print("----------------------------------------------------------------------")
else:
print("You have no passwords entered yet!")
while True:
chosen_option = logged()
print(chosen_option)
if chosen_option == "1":
viewapp()
elif chosen_option == "2":
addapp()
elif chosen_option == "3":
summary()
elif chosen_option == "4":
break
else:
print("That was not a valid option, please try again: ")
print("Goodbye")
发生这种情况是因为您在退出
addapp()
时调用了logged()
:然后,您输入的选择由
logged()
返回,并被丢弃,因为它没有分配给任何东西。在现在回到
addapp()
中前一个块的末尾,下一个指令是return addapp
,这将使您返回主循环,在那里,chosen_option = logged()
将再次将您发送到logged()
注意,在
return addapp
中,返回addapp
函数本身,这肯定不是您想要的。因此,由于您不需要addapp()
的返回值,只需使用return
,或者根本不使用,Python将在函数末尾自动返回。在所以,要解决您的问题:直接
^{pr2}$return
当您完成输入站点后:还请注意,当您添加更多站点时,您会从自身递归调用
addapp()
。除非您真的想使用某种递归算法,而是像在主循环中那样使用循环,否则通常应该避免这种情况。默认情况下,Python将你的递归级别限制在1000级,因此你甚至可以通过连续输入1000多个站点来崩溃应用程序;)
摘要问题仅由
summary()
中不必要的for
循环引起你快到了。问题出在第63行的addapp()函数中:
如果你更换
^{pr2}$与
那么一切都会好起来的。 无论如何,您不会在这里处理记录函数的结果。 您也不需要在这里处理记录的函数。addapp将退出,记录的函数将在调用addapp函数的while循环中被调用和处理。在
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