3

为学习目的构建一个简单的“石头、纸、剪刀”的 Python 游戏。

我在这里阅读了其他一些关于退出 Python 而没有回溯的帖子。我正在尝试实现它，但仍然得到追溯！一些 Python 奇才可以指出这个 Python 假人有什么问题吗？这个想法是单击 RETURN（或键入“yes”或“y”将使程序再次运行 play()，但按 CTRL-C 将关闭它而不会回溯。我使用的是 Python 2.7。

    # modules
    import sys, traceback
    from random import choice

    #set up our lists
    ROCK, PAPER, SCISSORS = 1, 2, 3
    names = 'ROCK', 'PAPER', 'SCISSORS'

    #Define a function for who beats who?
    def beats(a, b):
        return (a,b) in ((PAPER, ROCK), (SCISSORS, PAPER), (ROCK, SCISSORS))

    def play():
        print "Please select: "
        print "1 Rock"
        print "2 Paper"
        print "3 Scissors"
        # player choose Rock, Paper or Scissors
        player_choice = int(input ("Choose from 1-3: "))
        # assigns the CPU variable a random CHOICE from a list.
        cpu_choice = choice((ROCK, PAPER, SCISSORS))

        if cpu_choice != player_choice:
            if beats(player_choice, cpu_choice):
                print "You chose %r, and the CPU chose %r." % (names[player_choice - 1], names[cpu_choice - 1])
                print "You win, yay!!"
            else:
                print "You chose %r, and the CPU chose %r." % (names[player_choice - 1], names[cpu_choice - 1])
                print "You lose. Yuck!"
        else:
            print "You chose %r, and the CPU chose %r." % (names[player_choice - 1], names[cpu_choice - 1])
            print "It's a tie!"

        print "Do you want to play again? Click RETURN to play again, or CTRL-C to exit!"

        next = raw_input("> ")

        # THIS IS WHAT I'M WORKING ON - NEED TO REMOVE TRACEBACK!
        if next == "yes" or "y":
            try:
                play()
            except KeyboardInterrupt:
                print "Goodbye!"
            except Exception:
                traceback.print_exc(file=sys.stdout)
            sys.exit(0)
        elif next == None:
            play()
        else:
            sys.exit(0)

# initiate play() !
play()
4

3 回答 3

4

尝试重构你的主循环；更多类似的东西：

try:
    while (play()):
        pass
except KeyboardInterrupt:
    sys.exit(0)

play看起来像：

def play():
    _do_play() # code for the actual game

    play_again = raw_input('play again? ')
    return play_again.strip().lower() in ("yes", "y")
于 2012-09-17T20:09:46.347 回答
3

您调用play()了两次，因此您需要将两种情况都放在try/except块中：

if next in ("yes", "y"):
    try:
        play()
    except KeyboardInterrupt:
        print "Goodbye!"
    except Exception:
        traceback.print_exc(file=sys.stdout)
    sys.exit(0)
elif next is None:
    try:
        play()
    except KeyboardInterrupt:
        print "Goodbye!"
    except Exception:
        traceback.print_exc(file=sys.stdout)
        sys.exit(0)
else:
    sys.exit(0)

我已经纠正了另外两个问题，最好在 python中测试Nonewith ，而你的第一个测试不起作用，因为它被解释为与中间的 an分开。总是被视为因此您永远不会来到代码中的其他分支。isifnext == "yes" or "y"next == "yes""y"or"y"True

请注意，我怀疑上面的代码可以进一步简化，但是您根本没有向我们展示您的play()功能，所以您让我们猜测您要做什么。

于 2012-09-17T20:07:13.910 回答
2

一个问题是您需要将raw_input语句try except KeyboardInterrupt和实际play函数一起包含在子句中。例如

try:
   nxt = raw_input('>')
   if nxt.lower().startswith('y') or (nxt.strip() == ''):
      play()
   else:
      sys.exit(0)
except KeyboardInterrupt:
   sys.exit(0)
except Exception:
   traceback.print_exc(file=sys.stdout)
于 2012-09-17T20:03:48.590 回答