Question

我正在艰难地学习Python和学习Python，并且在 ex36 上遇到了问题。

我处于开发的早期阶段，我无法弄清楚我的if语句有什么问题。无论出于何种原因，我的代码永远不会过去

elif "1" or "2" in choice and not key.

即使声明中没有“1”或“2”。我不明白为什么会这样。看起来很好。当我使用另一个嵌套if语句时，嵌套语句超过了这一点，但是它挂起了另一个点，所以我移动了我的初始化变量 - 不确定这是否是python中的东西 - 我确实移动了它们虽然 - 在while循环之外。在我漫游之前，这里是完整的代码。我知道逻辑不完整，超过一半的代码没有完成，但我需要知道为什么这个语句不起作用。

#write function definition statements.
def darkRoom():
    door = raw_input(">>> ")

    if "1" in door:
        lions()
    elif "2" in door:
        tiger()
    elif "3" in door:
        bear()
    else:
        print """A thunderous voice booms through the room exclaiming,
"CHOOSE DOOR 1, 2, OR 3!"""
        darkRoom()    

def lions():
#naming error somewhere here
    keys = False
    lions = False #lions are calm if false. They are pissed if true
    warning = True
    while True:

        choice = raw_input(">>> ")
        print " %r %r %r" % (keys, lions, warning)
        x = "1" or "2" not in choice and not key and lions

        if "take" and "key" in choice:
            key = True
            print """There are two doors behind the angry pride of lions.
Which door are you going to run to and open before the lions eat you?"""
            door = raw_input(">>> ")
            if "1" in door and key == True:
                threeBrickRoads()
            elif "2" in door and key == True:
                quickSand()
            else:
                youDie("You take too long to decide which door to take and the lions eat you.")
        elif "look" in choice:
            print "Looks like you're going to have to take the key from the lions"
#never gets past this statement even when 1 and two not in choice. This is what my question
#is about
        elif "1" or "2" in choice and not key:
            print "The Door is locked and the lions are starting to stare."
            lions = True
            print " %r %r %r" % (keys, lions, warning)
            print "%r" % choice
#never reaches this point. I don't know why.

        elif x and warning:
                    print """The lions leave the key and start to chase you. Quick get the
key before they catch you"""

                    warning = False
#Statement never reaches this point. It should
        elif x and not warning:
                youDie("You take too long to take the key and the lions eat you for it.")
# entering jig in statement should put me here and not at line 46
        else:
            print """"You quickly realize that doesn't do you any good.
You take another look at your surroundings"""
#don't think I need that while I have the while loop.
        #lions()

##def tiger():

##def bear():

##def threeBrickRoads():

##def quickSand():

##def sizePuzzle():

##def riddlesOnWall():

##def wolfSheepCabbage():

##def duckHunt():

##def hangman():

##def goldRoom():

##def oceanShore():

##def winScreen():

def youDie():
    print why, """You lay there pondering your mistake as
the last fleeting pulses of life slowly beat out of you."""
    #exit(0)

darkRoom()

Answer 1

elif "1" or "2" in choice and not key

这解释如下（“1”或（（选择中的“2”）和（不是键）））

由于“1”始终为真，因此总是如此。我想你的意思是：

elif choice in ['1', '2'] and not key

Answer 2

让我们来看看这一行：

elif "1" or "2" in choice and not key:

这一行实际上表明它基本上要求以下两个条件之一为True：

如果“1”（没有其他任何内容）
如果选择“2”而不是键

如果您是初学者，这是一个典型的错误，如果您按照以下方式编写，则可以轻松解决此问题（最简单的修复）：

elif choice in [1, 2] and not key:

这意味着：如果选择等于列表中包含的任何元素[1,2]且密钥不是真的。

这个if语句有什么问题

3 个答案: