Question

我正在尝试完成我的程序我添加了一个菜单，允许用户选择一些允许用户在列表中存储网站名称和密码的选项。但是，只要我将一些网站名称和密码添加到各自的保险库中，当我在添加网站名称和密码后尝试选择一个选项时就会出现问题，＆＃34; 1＆＃34;例如，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，它拒绝第一个预期的输入，但奇怪地接受第二个输入。此外，当我选择第3个选项以调用summary（）时，整个打印的摘要将打印出两次，这与菜单仅接受第二个预期输入的类似模式。程序正在做我想要的事情，除了这个烦人的错误，选择这四个选项使得它第二次要求输入时，它应该立即跳转到该功能。帮助将不胜感激。

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")

Answer 1

这是因为您在退出logged()时致电addapp()：

if quit in ["end", "End", "END"]:
    logged()

然后，您输入的选项由logged()返回，并因未分配给任何内容而被丢弃。

您现在回到addapp()上一个块的末尾，下一条指令是return addapp，它将返回主循环，您将被发送到logged()再次chosen_option = logged()

请注意，在return addapp中，您会返回addapp 函数本身，这肯定不是您想要做的。因此，由于您不需要addapp()的返回值，只需使用return，或者根本不使用任何内容，Python将在函数结束时自动返回。

所以，要解决您的问题：当您完成输入网站时直接return：

if quit in ["end", "End", "END"]:
    return

另请注意，当您添加更多网站时，您会从自身递归调用addapp() 你应该完全避免这种情况，除非你真的想要使用一些递归算法，而是像你在主循环中那样使用循环。默认情况下，Python将您限制为1000个递归级别 - 因此您甚至可以通过连续输入1000多个站点来使应用程序崩溃;）

摘要问题仅由for

中不必要的summary()循环引起

Answer 2

你快到了。问题出在第63行的addapp（）函数中：

if quit not in ["end", "End", "END"]:
    logged()

如果你更换

logged()

与

pass

然后一切都会好起来的。无论如何，您无法在此处理已记录函数的结果。您也不需要在此处理记录的功能。 addapp将退出，并且将在调用addapp函数的while循环中调用和处理记录的函数。

Python：被要求输入两次并输出重复的打印语句

2 个答案: