Question

我正在尝试创建一个登录系统。我可以在不实现类而仅实现函数的情况下制作系统。我想将每个步骤都变成特定的方法，而不必全部写入一个函数中。

我的问题是，如果字符长度> 5或密码错误，如何恢复为登录名，询问用户名和密码。

如果用户名和密码不在列表中，我该如何将其还原或需要再次编码？

    class LoginSystem:

        def __init__(self):
            self.user_id = input("Please enter your user id: ")
            self.user_password = input("Please enter your password: ")

        def login(self):
            username = self.user_id
            password = self.user_password
            if len(username) <= 5 and len(password) <= 5:
                print("Logging In")
            else:
                print("Error! Max Length is 5 chars.") #return back to 
                login system

        def check_system(self):
            registered_user = {
             "test@gmail.com": "test"
            }
            if self.user_id in registered_user:
                print("Successful")
            else:
                new_user = input("Id not found! Are you are new user?\n [Y]es or [N]o\n")
                new_user = new_user.lower()
                if new_user == "Y":
                   return back to login system
                elif new_user == "N": #how to return back to main login system
                   new_username = input("Please enter your user id: ")
                   new_userpassword = input("Please enter your password: ")
                else:
                   return #back to login system

Answer 1

您的LoginSystem将错误的数据视为其实例属性。在各种方法调用之间，已注册用户的集合是不变的。 login方法本身应提示输入用户ID和密码。

class LoginSystem:

    def __init__(self):
        self.users = {"test@gmail.com": "test"}

    def login(self):
        while True:
            username = input("Please enter your user id: ")
            password = input("Please enter your password: ")
            if len(username) <= 5 and len(password) <= 5 and self.check_system(username, password):
                print("Logging In")
                break
            else:
                # TODO Disallow infinite retries to get it right
                print("Error! Max Length is 5 chars.")

    def check_system(self, name, password):
        try:
            expected_password = self.registered_user[name]
        except KeyError:
            # Or try to add a new user to the system here
            return False

        if password != expected_password:
            return False

        return True

可以添加一个单独的方法，以便在必要时向系统添加新用户。

创建登录系统

1 个答案: