Question

def flips():
        print "Alright the workout of the day is a front-flip, back-flip and
                       side flip"                       
        flips_left = ['side flip', 'front flip', 'back flip']
        flip_choice = None

        while len(flips_left) > 0:
            flip_choice = raw_input("Do a flip ")
            if "side flip" in flip_choice:
                 print flip_lines['side_flip']
                 flips_left.remove('side flip')
                 print "Great! Now just do the rest: %s" % flips_left
             elif "front flip" in flip_choice:
                 print flip_lines['front_flip']
                 flips_left.remove('front flip')
                 print "Good stuff! This is what you have left: %s" %
                                 flips_left
             elif "back flip" in flip_choice:
                 print flip_lines['back_flip']
                 flips_left.remove('back flip')
             else:
              "That is not a type of flip! Try again!"

         print "Great! You completed the WOD!"

这个问题实际上是两个部分：

我正在做的是告诉用户输入'侧翻'，'后翻'或'前翻'。当他们这样做时，输入将从列表'flips_left'中删除。这样工作正常，直到说用户再次进入相同的翻转，我想返回一条消息，告诉他们他们已经这样做了，并再试一次。
由于某些原因，当用户输入的内容不是“侧翻”，“正面翻转”或“后翻”时，我无法让程序打印else语句。有任何想法吗？谢谢！

Answer 1

您的if语句逻辑未检查flips_left

中剩余的内容

您最好的选择是做一些更通用的事情，例如：

if flip_choice in flips_left:
    print flip_lines[flip_choice]
    flips_left.remove(flip_choice)
else:
    print "Some error"

如果您想跟踪已经做出的选择，可以使用字典来跟踪已经执行的翻转。

flips_left = {'side flip':True, 'front flip':True, 'back flip':True}

...

if flip_choice in flips_left:
    if flips_left[flip_choice]:
        print flip_lines[flip_choice]
        flips_left[flip_choice] = False
    else:
        print "You have already done %s!" % flip_choice
else:
    print "That is not a valid flip"

Answer 2

尝试在邮件前添加print关键字：

else:
    print "That is not a type of flip! Try again!"

列表为空时打印字符串而不是错误

2 个答案: