在 python 中玩猜数游戏，直到用户选择退出

Question

import random

play_game=input("Do you want to play ? Y/N :")
play_game=play_game.upper()
print(play_game)

while play_game == 'Y':
    #Type of game does the user wish to play
    game_type=input("please select a type:'H' for hard,'M' for medium ,'E' for easy  :")
    game_type=game_type.upper()

    #print(game_type)
    if game_type=='E':
        #generating a random number between 1 and 10
        random_number=random.randint(1,10)
        print(random_number)

        guess_times=0
        #guessing a number
        while guess_times<3:
            user_number=(input("Enter the number \n:"))
            user_number=int(user_number)
            guess_times+=1
            if user_number!=abs(user_number):
                print("the number should be a +ve number")
            else:
                if user_number==random_number:
                    print("Congratulaions,%d is the correct guess and you guessed in %d attempts"%(user_number,guess_times))
                    break
                if user_number<random_number:
                    print("the number,%d is lower than the guessed number"%user_number)
                if user_number>random_number:
                    print("the number,%d is higher than the guessed number"%user_number)
else:
        print("not the valid option")
print("bye")

我在上面创建了一个简单的猜谜游戏。它工作正常。但我希望用户玩游戏直到用户决定不玩。

我在游戏开始时使用 while 循环，一旦用户进入游戏，就不会选择退出选项。

我还不想使用函数，因为我是编程新手。

Answer 1

您永远不会在循环中更新 play_game；您需要在循环结束时再次询问用户：

while play_game == 'Y':
    # your game 

    play_game = input("Do you want to play again? Y/N :").upper()

然后，当循环回到顶部时，如果用户没有在提示时输入 y 或 Y，则游戏结束。

Answer 2

您可以为 game_type 添加一个选项，例如 Q 表示退出：

game_type=input("please select a type:'H' for hard,'M' for medium ,'E' for easy , Q to quit :")

并像这样更改代码的末尾：

elif game_type == 'Q':
    print("bye")
    break
else:
    print("not the valid option")

在 python 中玩猜数游戏，直到用户选择退出

playing a guess number game in python till the user opts to quit

python

python-3.x