我将如何定义名称验证

问题描述

我想做什么：

我正在尝试定义一个接受输入（名称作为字符串）并检查名称是否在设置列表中的定义。

如果是，该程序将继续。 如果不是，它将为用户提供两次输入正确名称的机会。

要求：

我需要在定义后使用name变量。 如果名称错误3次，我需要退出系统。

问题：

如果名称正确，则可以正常工作 但是，如果名称错误，则不允许输入其他名称，打印“您还有2次尝试”和“您还有1次尝试”，然后结束循环并退出。

代码：

vector

解决方法

您的代码有一些问题：

• 您永远不会在循环中要求用户输入新内容，而只是再次测试名字
• 您修改（本地）name_1变量，但不要return修改给调用者的值
• 您不必重复所有条件，可以使用数学来确定剩余的尝试次数

您可以尝试以下操作：

def get_name(player,tries,names_allowed_to_play):
    print(f"Player {player},you have {tries} tries to enter the correct name")
    for i in range(1,tries+1):
        name = input("Please enter your name to play: ").upper()
        if name in names_allowed_to_play:
            print("You are authorised to play,have fun!")
            names_allowed_to_play.remove(name)
            return name
        elif i < tries:
            print(f"You have {tries - i} more tries")
        else:
            print("You messed up")
            exit()

names_allowed_to_play = ["MUM","DAD"]
name1 = get_name(1,3,names_allowed_to_play)
name2 = get_name(2,names_allowed_to_play)
print(name1,name2)

,

这对我有用，我认为您的结构方式意味着逻辑存在缺陷。

import sys,time

names_allowed_to_play = ["MUM","DAD"]

def main():
    authorised = False
    attempts = 3
    while authorised == False:
        if attempts < 3:
            print("You have",attempts,"remaining.")
        if attempts > 0:
            name = input("Player 1,please enter your name to play: ")
            name = name.upper()
        elif attempts <= 0:
            print("You are locked out.")
            time.sleep(1)
            sys.exit()
        if name in names_allowed_to_play:
            print(name," you are authorised to play,have fun!")
            names_allowed_to_play.remove(name)
            authorised = True
        else:
            attempts -= 1
main()

,

主要更改：

• 将 input（）放入while循环，以便在名称错误时可以获取新输入，否则只能获取一次输入，并且if语句会检查错误的名称3次
• 为允许的名称添加 return 语句，以便稍后使用该名称（例如打印输出或其他功能）

您可以在我的代码注释中找到其他小的更改。

# change the names to lowercase so you don't need upper() for input
names_allowed_to_play = ["mum","dad"]

def player_validate():
    # \n for new line,just works like your print("")
    print("Player 1,you have 3 tries to enter the correct name\n")
    tries = 0

    while tries < 3:
        player_name = input("Player 1,please enter your name to play: ")
        if player_name in names_allowed_to_play:
            print(player_name + ",you are authorised to play,have fun!\n")
            names_allowed_to_play.remove(player_name)
            # return will stop the loop so you don't need to break it by other code
            return player_name
        else:
            tries = tries + 1
            # use calculation instead of many elif statment
            print(f"You have {3-tries} tries.")

    # I have change Player 2 to Player 1,if it is not typo,you can change it back
    print("Sorry Player 1," + player_name + " ruined it! " + player_name + ",you are NOT AUTHORISED!")
    # use exit() instead of sys.exit() or you will need to import sys at beginning
    exit()

# Run function (I think function is a more common name then definition)
name_to_use_later = player_validate()
print(name_to_use_later)

,

我已经解决了您的代码问题。

names_allowed_to_play = ["MUM","DAD"]
def val_1():
   print("Player 1,you have 3 tries to enter the correct name")
   name_1 = input("enter your name")
   a=0
   while a < 3:
       name_1 = name_1.upper()
       if name_1 in names_allowed_to_play:
           print(name_1 + ",have fun!")
           a = a + 4
           names_allowed_to_play.remove(name_1)
       elif name_1 not in names_allowed_to_play:
           a = a + 1
           if name_1 not in names_allowed_to_play and a == 1:
               print("You have 2 more tries")
               name_1 = input("enter your name")
               print("")
           elif name_1 not in names_allowed_to_play and a == 2:
               print("You have 1 more try")
               name_1 = input("enter your name")
           elifa == 3:
               print("")
               print("Sorry Player 2," + name_1 + " ruined it! " + name_1 + ",you are NOT AUTHORISED!")
               exit()

val_1()