[英]Python 3.X jumping to a line of code
我正在尝试使用Python 3.X创建一个“选择自己的冒险”游戏。 我正在使用空闲平台。 这是我的代码示例。
import time
again = True
while again == True:
print("You wake up as you normally do, but something's not quite right. What will be the first thing you do? \n Choices: \n Go back to sleep \n Get up")
action1 = input(": ")
if action1 == "Go back to sleep":
print("You lazyhead. As you were about to fall asleep, your ceiling collapsed on you and killed you. Don't be lazy.")
playAgain = input("Play again? Y/N: ")
if playAgain == "Y":
again = True
elif playAgain == "N":
again = False
显然,在action1
和playAgain
之间有东西。 我想在elif playAgain =="N"
again = false
替换again = false
elif playAgain =="N":
again = false
我想跳playAgain
再次playAgain
所以我不只是结束程序。 这是我的第一个问题,因此我的格式可能有点怪异,因此也请更正。
如前所述,python不支持GoTo
。 实现相同结果的一种方法是创建一个退出游戏函数,如下所示,然后在主游戏代码的末尾调用它。
def quit_game():
playAgain = input("Play again? Y/N: ")
if playAgain == "Y":
return True
elif playAgain == "N":
quit_game()
用原始代码:
...
again = quit_game()
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