如果 function 内部发生了应该停止迭代的事情，我该如何停止 while 循环？

Question

各位程序员，您好，我是 python 的初学者，几个月前。 我决定开始自己的小项目，以帮助我了解 Python 的整个开发过程。 我简要了解了所有基本语法，但我想知道如何在 function 调用 while 循环结束时进行一些操作。

我正在创建一个简单的终端号码猜谜游戏，它由玩家多次尝试猜测 1 到 10 之间的数字（我目前将其设置为 1 以测试代码中的某些内容）。

如果玩家得到正确的数字，则该级别应该结束，然后玩家将进入下一个级别的游戏。 我试图创建一个变量并做出一个真正的错误陈述，但我无法在 while 循环内操作 function 中的变量。

我想知道如何才能使游戏在玩家获得正确的数字时结束，我将在此处包含我的代码，以便你们有更多的上下文：

import random
import numpy
import time

def get_name(time):

    name = input("Before we start, what is your name? ")

    time.sleep(2)
    print("You said your name was: " + name)


# The Variable 'tries' is the indication of how many tries you have left
tries = 1

while tries < 6:

    def try_again(get_number, random, time):

        # This is to ask the player to try again
        answer = (input(" Do you want to try again?"))
        time.sleep(2)

        if answer == "yes":

            print("Alright!, well I am going to guess that you want to play again")
            time.sleep(1)

            print("You have used up: " + str(tries) + " Of your tries. Remember, when you use 5 tries without getting the correct number, the game ends")

        else:
            print("Thank you for playing the game, I hope you have better luck next time")
    

    def find_rand_num(get_number, random, time):

        num_list = [1,1]
        number = random.choice(num_list)

        # Asks the player for the number
        ques = (input("guess your number, since this is the first level you need to choose a number between 1 and 10  "))
        print(ques)

        if ques == str(number):
            time.sleep(2)

            print("Congratulations! You got the number correct!")
            try_again(get_number, random, time)


        elif input != number:
            time.sleep(2)

            print("Oops, you got the number wrong")
            try_again(get_number, random, time)
            
    def get_number(random, try_again, find_rand_num, time):
        
        # This chooses the number that the player will have to guess  
        time.sleep(3)              
        print("The computer is choosing a random number between 1 and 10... beep beep boop")
        time.sleep(2)

        find_rand_num(get_number, random, time)


    if tries < 2:
        get_name(time)
        tries += 1
        get_number(random, try_again, find_rand_num, time)

    else:
        tries += 1
        get_number(random, try_again, find_rand_num, time)


    if tries > 5:
        break

对于代码中的一些格式，我深表歉意，我尽我所能看起来和我的 IDE 中的一样准确。 我父亲通常会帮助我解决这些类型的问题，但在这一点上，我似乎比我父亲了解更多 python，因为他从事前端 web 开发。 所以，回到我原来的问题，如果这个声明：

    if ques == str(number):
        time.sleep(2)

        print("Congratulations! You got the number correct!")
        try_again(get_number, random, time)

是真的，while 循环结束了吗？ 另外，我的代码看起来如何？ 我花了一些时间使它看起来整洁，从专家的角度来看我很感兴趣。 我曾经在编程中读到过，少即是多，所以我试图用我的代码少花钱多办事。

感谢您花时间阅读本文，如果你们中的某些人对我的问题有任何解决方案，我将不胜感激。 祝你有美好的一天！

Answer 1

您的代码中有太多错误。 首先，您从未使用过在函数中传递的参数，所以我看不出有理由让它们留在那里。 然后你需要从你的函数中返回一些东西来使用它们来打破条件（例如True/False ）。 最后，我想在您的情况下单独调用函数要方便得多，因为您需要在继续之前进行一些检查（不在彼此内部）。 所以，这是我最终得到的代码：

import random
import time

def get_name():
    name = input("Before we start, what is your name? ")

    time.sleep(2)
    print("You said your name was: " + name)


def try_again():
    answer = (input("Do you want to try again? "))
    time.sleep(2)

    # Added return True/False to check whether user wants to play again or not
    if answer == "yes":
        print("Alright!, well I am going to guess that you want to play again")
        time.sleep(1)

        print("You have used up: " + str(tries) + " Of your tries. Remember, when you use 5 tries without getting the correct number, the game ends")
        return True

    else:
        print("Thank you for playing the game, I hope you have better luck next time")
        return False


# Joined get_number and find_random_number since get_number was doing nothing than calling find_rand_num
def find_rand_num():
    time.sleep(3)              
    print("The computer is choosing a random number between 1 and 10... beep beep boop")
    time.sleep(2)

    num_list = [1,1]
    number = random.choice(num_list)

    ques = (input("guess your number, since this is the first level you need to choose a number between 1 and 10  "))
    print(ques)

    if ques == str(number):
        time.sleep(2)
        print("Congratulations! You got the number correct!")
        # Added return to check if correct answer is found or not
        return "Found"
    elif input != number:
        time.sleep(2)
        print("Oops, you got the number wrong")


tries = 1
while tries < 6:
    if tries < 2:
        get_name()
    
    res = find_rand_num()
    if res == "Found":
        break

    checker = try_again()
    if checker is False:
        break
    
    # Removed redundant if/break since while will do it itself
    tries += 1