Python 我怎么能在一个输入中做一个有限的数字，并且只做数字，我不想要符号

我有一个小游戏，它让你选择一个随机的游戏，但我有两个问题：

1-我想验证用户输入是否为介于0和20之间的数字

2-我想让用户再次播放（可以通过询问他或直接播放）

这是我的代码，如果有任何问题或建议，请告诉我，我希望你能写在一封信中，这样我就可以理解它，非常感谢你的帮助

代码如下：

import random
rn = int(random.randint(0, 20))
gn = int(input("write a number from 0-20: "))


while True:
  if rn < gn:
    print("You were too close, try again")
    break
  elif rn == gn:
    print("Oh, I lost, you win, good game (GG)")
    break
  if rn > gn:
    print("You are not close, try again")
    break

while False:
  rn = int(random.randint(0, 20))
gn = int(input("write a number from 0-20: "))

随机导入
rn=int（random.randint（0,20））
gn=int（输入（“写一个0-20之间的数字：”）
尽管如此：
如果rngn：
打印（“您尚未关闭，请重试”）
打破
虽然错误：
rn=int（random.randint（0,20））
gn=int（输入（“写一个0-20之间的数字：”）

我将代码尽可能模块化，并且非常简单

def checkInput(inp):
    try:
        n = int(inp)
    except:
        return (False,"Please enter an integer between 0 to 20 and nothing else")
    if n >=0 and n<=20:
        return (True,n)
    else:
        return (False, "Your number is not in the expected range of 0 to 20")

def gameInput():
    import random
    rn = int(random.randint(0, 20))
    gn = input("write a number from 0-20: ")

    while(not checkInput(gn)[0]):
        print(checkInput(gn)[1])
        gn = input("write a number from 0-20: ")
    gn = checkInput(gn)[1]
    return gn,rn

def gameCheck():
    gn,rn = gameInput()
    # print(gn,rn)
    if rn < gn:
        print("You were too close, try again")
    elif rn == gn:
        print("Oh, I lost, you win, good game (GG)")
    elif rn > gn:
        print("You are not close, try again")


def game():
    gameCheck()
    playAgain = input("Do you want to play again? Enter Y for yes and anything else for no")
    while playAgain.lower() =="y":
        gameCheck()
        playAgain = input("Do you want to play again? Enter Y for yes and anything else for no")

game()

def检查输入（inp）：
尝试：
n=int（inp）
除：
return（False，“请输入一个介于0到20之间的整数，不要输入其他值”）
如果n>=0且n gn:
打印（“您尚未关闭，请重试”）
def game（）：
gameCheck（）
再次播放=输入（“您想再次播放吗？输入Y表示是，输入其他任何内容表示否”）
再次播放时。lower（）=“y”：
gameCheck（）
再次播放=输入（“您想再次播放吗？输入Y表示是，输入其他任何内容表示否”）
游戏（）

要检查输入值，请尝试以下操作：

def ask_for_guess():
    # the function will ask infinitely for input number unless it's valid number in between 0-20 
    while True:
        guess_input = input("write a number from 0-20: ")
        try:
            gn = int(guess_input) 
            assert 0 <= gn <= 20, "Number must be between 0-20"
            return gn
        except ValueError as ex:
            # ValueError will be thrown if the conversion to int will fail
            print(f"{guess_input} is not a number, please try again")
        except AssertionError as ex:
            # AssertionError will be thrown by the test in the try block if the number is not in range
            print(str(ex))

def ask_for_guess（）：
#函数将无限次地请求输入数字，除非它是介于0-20之间的有效数字
尽管如此：
猜一猜输入=输入（“写一个0-20之间的数字：”）
尝试：
gn=int（猜测输入）
断言0您将首先重新阅读if和while语句的工作方式。你的循环是非感官的。
play_again = True
while play_again: 
    rn = int(random.randint(0, 20))
    gn = ask_for_guess()
    if rn < gn:
        print("You were too close, try again")
    elif rn == gn:
        print("Oh, I lost, you win, good game (GG)")
    else:
        print("You are not close, try again")
    # if the user doesn't want to play again, the play_again variable will be False and the code won't enter for another iteration inside the while loop
    play_again = input("Do you want to play again? [y/n]") == "y"