猜猜python中的数字。结束游戏的选项

我是Python和编程新手，我正在尝试制作“猜数字”游戏。我让它工作，但我想添加选项，以关闭游戏使用刹车声明。我已经试了两天了，但我想我卡住了

以下是有效的代码（没有关闭游戏的选项）：

随机导入
尝试次数=0
secretNumber=random.randint（1100）
尽管如此：
打印（“猜测1到100之间的数字”）
猜测=输入（）
猜测=int（猜测）
尝试次数=尝试次数+1
如果猜测>100或猜测0：
打印（“不！请尝试更高的数字！\n”）
elif guess>secretNumber和guess 100或guess 0：
打印（“不！请尝试更高的数字！\n”）
elif guess>secretNumber和guess变化不大，一切正常

import random
attempts = 0
secretNumber = random.randint(1,100)
while True:
    print("Guess a number between 1 and 100. You can exit game anytime by typing q ")
    guess=input()
    stop=str(guess)
    if stop== 'q':
        print("You quitted the game")
        break
    elif not guess.isdigit():
        print("Only numbers are allowed")
    else:        
        guess = int(guess)
        attempts = attempts + 1
        if guess >100 or guess<=0:
            attempts = attempts - 1
            print("Not allowed.Only numbers between 1 and 100 allowed! \n")
        elif guess < secretNumber and guess >0:
            print("Nope!Try a higher number! \n")
        elif guess > secretNumber and guess <100:
            print("Nope!Try a lower number \n")
        if guess == secretNumber:
            attempts = str(attempts)
            print("Congratulations you found it after " + attempts + " attempts")
            break  

随机导入
尝试次数=0
secretNumber=random.randint（1100）
尽管如此：
打印（“猜一个介于1和100之间的数字。您可以通过键入q随时退出游戏”）
猜测=输入（）
停止=str（猜测）
如果停止=='q'：
打印（“你退出了游戏”）
打破
elif not guess.isdigit（）：
打印（“只允许数字”）
其他：
猜测=int（猜测）
尝试次数=尝试次数+1
如果猜测>100或猜测0：
打印（“不！请尝试更高的数字！\n”）
elif guess>secretNumber和guess在Nick A和timgeb在评论中提到的内容下面，您可以修改一些内容。
正如我在下面的脚本中所建议的那样，您可能想看看。请在脚本中查找注释

import random
attempts = 0
secretNumber = random.randint(1,100)
# you can prompt the user within input()
stop=input("Guess a number between 1 and 100. You can exit game anytime by typing quit ")

while True:
    if stop == "quit": # as the the output of input() is a string, you can compare it with a string without str()
        print("You quitted the game")
        break
    try: # work with try and except to avoid problems whenever the user writes something else than 'quit' or an integer
        guess = int(stop)
        attempts += 1
        if guess >100 or guess<=0:
            attempts = attempts - 1
            print("Not allowed.Only numbers between 1 and 100 allowed! \n")
        elif guess < secretNumber and guess >0:
            print("Nope!Try a higher number! \n")
        elif guess > secretNumber and guess <=100: # note that it's <=. Otherwise nothing happens when the users chose 100
            print("Nope!Try a lower number \n")
    except: # if it's not possible to turn 'stop' into an integer we'd like to perform the following
        stop = input("No valid input, please guess again ") # Don't stop the script, let the user guess again
        continue # don't stop, but leave the current iteration (don't execute the rest) and start again at the beginning of the loop 
    if guess == secretNumber:
        attempts = str(attempts)
        print("Congratulations you found it after " + attempts + " attempts")
        break
    stop = input("Guess again ") # prompt for a new number every round of the loop

随机导入
尝试次数=0
secretNumber=random.randint（1100）
#您可以在input（）中提示用户
停止=输入（“猜一个介于1和100之间的数字。您可以通过键入quit随时退出游戏”）
尽管如此：
如果stop==“quit”：#由于input（）的输出是一个字符串，因此可以将其与不带str（）的字符串进行比较
打印（“你退出了游戏”）
打破
try:#使用try和except，以避免在用户写入“quit”或整数以外的内容时出现问题
猜测=int（停止）
尝试次数+=1
如果猜测>100或猜测0：
打印（“不！请尝试更高的数字！\n”）
elif guess>secretNumber和guess删除您的
否则：继续
<代码>继续
使代码在循环开始时再次继续。并且在中断之前打印
。@Nick A和timgeb感谢您的回答。我两个都做了，但仍然不起作用。输入一个数字后，什么也没有发生，我必须再次键入该数字才能工作，如果我不这样做，只需再次按enter键，我会得到一个ValueError，当我键入quit时，它会工作，但我会得到一个NameError..非常感谢@legendisback！！绝对是一个传奇……它成功了！我一直在尝试使用.isdigit（）方法，以避免在用户输入str而不是int时出现ValueError，我尝试将“q”字更改为0，以避免在没有运气的情况下发生冲突。这可能吗？有什么想法吗？非常感谢您的时间。@Mike\u R现在编辑了这个程序。添加了isdigit（）只允许数字猜测！再次非常感谢！在Python3中，
input
中的值已经是一个字符串，无需使用
str（）
进行转换，它工作得很好，但我不希望当用户提交无效值时程序停止，因此我删除了“except:”后面的“break”，循环不会停止。当第一个提交的值无效时就会出现问题..我得到一个NameError。有没有想过我该如何解决这个问题？再次感谢您抽出时间！如果您删除了
break
，并且用户提交了一个无效输入，则不会定义
guess
，因为无法将无效输入转换为整数。基本上，您可以更改相应的
if
语句，并使用
continue
而不是
break
。我对脚本进行了更新，并对其工作原理进行了评论。您可以了解有关continue语句的更多信息。请随时提出任何进一步的问题！
import random
attempts = 0
secretNumber = random.randint(1,100)
while True:
    print("Guess a number between 1 and 100. You can exit game anytime by typing q ")
    guess=input()
    stop=str(guess)
    if stop== 'q':
        print("You quitted the game")
        break
    elif not guess.isdigit():
        print("Only numbers are allowed")
    else:        
        guess = int(guess)
        attempts = attempts + 1
        if guess >100 or guess<=0:
            attempts = attempts - 1
            print("Not allowed.Only numbers between 1 and 100 allowed! \n")
        elif guess < secretNumber and guess >0:
            print("Nope!Try a higher number! \n")
        elif guess > secretNumber and guess <100:
            print("Nope!Try a lower number \n")
        if guess == secretNumber:
            attempts = str(attempts)
            print("Congratulations you found it after " + attempts + " attempts")
            break  

import random
attempts = 0
secretNumber = random.randint(1,100)
# you can prompt the user within input()
stop=input("Guess a number between 1 and 100. You can exit game anytime by typing quit ")

while True:
    if stop == "quit": # as the the output of input() is a string, you can compare it with a string without str()
        print("You quitted the game")
        break
    try: # work with try and except to avoid problems whenever the user writes something else than 'quit' or an integer
        guess = int(stop)
        attempts += 1
        if guess >100 or guess<=0:
            attempts = attempts - 1
            print("Not allowed.Only numbers between 1 and 100 allowed! \n")
        elif guess < secretNumber and guess >0:
            print("Nope!Try a higher number! \n")
        elif guess > secretNumber and guess <=100: # note that it's <=. Otherwise nothing happens when the users chose 100
            print("Nope!Try a lower number \n")
    except: # if it's not possible to turn 'stop' into an integer we'd like to perform the following
        stop = input("No valid input, please guess again ") # Don't stop the script, let the user guess again
        continue # don't stop, but leave the current iteration (don't execute the rest) and start again at the beginning of the loop 
    if guess == secretNumber:
        attempts = str(attempts)
        print("Congratulations you found it after " + attempts + " attempts")
        break
    stop = input("Guess again ") # prompt for a new number every round of the loop