Question

我正在练习python挑战。在这里，我尝试制作一个简单的数字猜谜游戏，您可以随时输入exit而不是猜测来退出游戏。

此刻我的代码就在这里。

import random
def run():
number = random.randint(1,9)
print ("Hi, your mission is to guess the secret number. You\'ll get a clue, it\'s a number between 1 and 9 (including 1 and 9). \n Type exit whenever to exit game.")
guess = ""
guesses = 0
while guess != number:
    guess = (input("Guess a number: "))
    guesses += 1
    if int(guess) < number:
        print ("Too low...")
    elif int(guess) > number: print("Too high!")
    elif str(guess) == ("exit"):
        return "Thank you for playing!"
    else:
        print ("Good job! You guessed the secret word in {} number of tries!".format(guesses))
        run()
run()

好！因此，无效的部分是“ elif str（guess）==（“ exit”）：返回“谢谢您的游戏！”

编辑：

import random
def run():
while True:
    number = random.randint(1,9)
    print ("Hi, your mission is to guess the secret number. You\'ll get a clue, it\'s a number between 1 and 9 (including 1 and 9). \n Type exit whenever to exit game.")
    guess = ""
    guesses = 0
    while guess != number:
        guess = (input("Guess a number: "))
        guesses += 1
        if type(guess) is int and guess < number:
            print ("Too low...")
        elif type(guess) is int and guess > number: print("Too high!")
        elif type(guess) is str and guess == ("exit"):
            return "Thank you for playing!"
        else:
            print ("Good job! You guessed the secret word in {} number of tries!".format(guesses))
            break

run（）

但这只给了我else陈述？（“干得好！你。。。”）

通过重写解决，并在解决方案中获得了一些帮助，并意识到整个订单被弄乱了。新的工作代码：

print("Hi, your mission is to guess the secret number. You\'ll get a clue, it\'s a number between 1 and 9 (including 1 and 9). \nType exit whenever to exit game.")
import random
number = random.randint(1, 9)
guess = 0
count = 0
    while guess != number and guess != "exit":
    guess = input("\nWhat's your guess? ")
    if guess == "exit":
        print ("Thank you for playing!")
        break
    guess = int(guess)
    count += 1
    if guess < number:
        print("Too low!")
    elif guess > number:
        print("Too high!")
    else:
        print("You got it!")
        print("And it only took you", count, "tries!")

Answer 1

您正在对int()的值进行str()和guess类型转换，然后在不首先检查实际数据类型的情况下进行比较。这绝对不是一件好事，因为一旦您输入“ exit”（或其他任何str，float或其他无效类型），程序将抛出ValueError。

您可以按照以下方法检查guess的类型，而不是进行类型转换。因为我们已经知道guess的类型就是程序所期望的，所以这不需要类型转换。

if type(guess) is int and guess < number:
    print ("Too low...")
elif type(guess) is int and guess > number: 
    print("Too high!")
elif type(guess) is str and guess == ("exit"):
    return "Thank you for playing!"

Answer 2

当用户正确猜出一个数字时，您正在进行递归调用，因此即使用户在下一个游戏中输入“退出”，它也只会返回到调用run，而不是主程序。您应该使用外部while循环来使游戏继续进行，直到用户正确猜出一个数字：

import random
def run():
    while True:
        number = random.randint(1,9)
        print ("Hi, your mission is to guess the secret number. You\'ll get a clue, it\'s a number between 1 and 9 (including 1 and 9). \n Type exit whenever to exit game.")
        guess = ""
        guesses = 0
        while int(guess) != number:
            guess = (input("Guess a number: "))
            guesses += 1
            if guess == "exit":
                return "Thank you for playing!"
            elif int(guess) < number:
                print ("Too low...")
            elif int(guess) > number:
                print("Too high!")
            else:
                print ("Good job! You guessed the secret word in {} number of tries!".format(guesses))
run()

Python 3-在猜游戏中随时键入“ exit”退出程序

2 个答案: