python - 我在哪里檢查我的石頭剪刀布游戲中的有效輸入

Question

當用戶輸入'r'，'p'，'s'之間的選擇時，我想檢查有效輸入我將如何 go 這樣做？ 我把它放在我的代碼的什么地方？

我已經嘗試在開始時添加它，但它沒有用：

possible_choices = ["r", "p", "s"]

def play():
    user = input(
        "What's your choice? 'r' for rock, 'p' for paper, 's' for scissors: "
    ).lower
    computer = random.choice(["r", "p", "s"])
    if user not in possible_choices:
        print('Uh oh! Please select a valid option', play())

整個代碼：

import random

game_loop = True


def play():
    user = input(
        "What's your choice? 'r' for rock, 'p' for paper, 's' for scissors: "
    ).lower
    computer = random.choice(["r", "p", "s"])

    if user == computer:
        return f"Tie! Computer chose {computer}"

    # r > s, s > p, p > r
    if win_or_lose(user, computer):
        return f"You won! Computer chose {computer}"
    return f"You lost! Computer chose {computer}"


def win_or_lose(player, opponent):

    if (
        (player == "r" and opponent == "s")
        or (player == "s" and opponent == "p")
        or (player == "p" and opponent == "r")
    ):
        return True


def play_again():
    p_a = input("Play again? (y/n): ")
    if p_a == "y":
        print(play())
    elif p_a == "n":
        quit()
    else:
        print("oof! Please choose correct option: (y/n)")
        play_again()


print(play())

while game_loop == True:
    play_again()

Answer 1

我推薦一個while循環，實現如下：

def play():
    choices = ['r','p','s']
    while True:
        user = input(
            "What's your choice? 'r' for rock, 'p' for paper, 's' for scissors: "
        ).lower()
        if user in choices:
            break
        else:
            print('choose again')
            continue
...

這個結構檢查輸入是否在選項列表中，如果不是，用戶再試一次。 還添加了 the.lower() 上的括號。

可以在此處找到有關此類事物的更多信息

Answer 2

回答：

def play():
    choices = ["r", "p", "s"]
    computer = random.choice(["r", "p", "s"])
    while True:
        user = input(
            "What's your choice? 'r' for rock, 'p' for paper, 's' for scissors: "
        )
        if user in choices:
            break
        else:
            print("Please choose a correct choice.")
            continue

    if user == computer:
        return f"Tie! Computer chose {computer}"

    # r > s, s > p, p > r
    if win_or_lose(user, computer):
        return f"You won! Computer chose {computer}"
    return f"You lost! Computer chose {computer}"


def win_or_lose(player, opponent):

    if (
        (player == "r" and opponent == "s")
        or (player == "s" and opponent == "p")
        or (player == "p" and opponent == "r")
    ):
        return True


def play_again():
    while True:
        p_a = input("Play again? (y/n): ")
        if p_a == "y":
            print(play())
        elif p_a == "n":
            quit()
        else:
            print("Please choose correct option..")
        play_again()


print(play())

while True:
    play_again()