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IndentationError:如果player['health'] <= 0:
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Stack Overflow用户
提问于 2019-10-18 10:02:57
回答 1查看 38关注 0票数 0

我在这个代码中的错误在哪里仍然是个乞丐

代码语言:javascript
复制
player = {'name': 'Frankie', 'attack': 10, 'heal': 5, 'health': 200}
eliza = {'name': 'Eliza', 'attack': 5, 'health': 200}
game_running = True

while game_running == True:

    print('please select action')
    print('1) Attack')
    print('2) Heal')

    player_choice = input()

    if player_choice == '1':
        eliza['health'] = eliza['health'] - player['attack']
        player['health'] = player['health'] - eliza['attack']
        print(eliza['health'])
        print(player['health'])


    elif player_choice == "2":
        print('Heal player')
    else:
        print('Invalid Input')

     if player['health'] <= 0:
         game_running = False

我的问题是,如果玩家‘健康’<= 0:我错在哪里?

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回答 1

Stack Overflow用户

回答已采纳

发布于 2019-10-18 10:21:18

代码语言:javascript
复制
player = {'name': 'Frankie', 'attack': 10, 'heal': 5, 'health': 200}
eliza = {'name': 'Eliza', 'attack': 5, 'health': 200}
game_running = True

while game_running == True:

    print('please select action')
    print('1) Attack')
    print('2) Heal')

    player_choice = input()

    if player_choice == '1':
        eliza['health'] = eliza['health'] - player['attack']
        player['health'] = player['health'] - eliza['attack']
        print(eliza['health'])
        print(player['health'])


    elif player_choice == "2":
        print('Heal player')
    else:
        print('Invalid Input')

    if player['health'] <= 0:
        game_running = False

这里的代码运行,我在本地测试了它。

建议:

indenting.

  • while game_running

  • 使用PyCharm或其他相关的IDE,这样您就可以在运行脚本之前习惯并进行概述(在PyCharm中,您将得到一条红线,告诉您while game_running == True

有问题就足够了,您不必显式地编写while game_running == True)。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/58448525

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