这个if语句有什么问题

Question

我在自学python，学习Python很困难，我遇到了一个问题ex36。

我处于开发的相当早期阶段，我无法弄清楚我的if语句有什么问题。不管出于什么原因，我的代码永远不会使得过去

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

即使“1”或“2”都没有在声明。我不明白为什么会发生这种情况。看起来很好。当我为此使用另一个嵌套的if语句时，嵌套语句通过了这一点，但它被挂在另一个点上，所以我移动了我的初始化变量 - 不太确定这是否是python中的事情 - 我确实移动了它们尽管 - 在while循环之外。 在我漫不经心的时候，下面是代码的全部内容。 我知道逻辑不完整，并且超过一半的代码没有完成，但我需要知道为什么这个声明不起作用。

#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

让我们看看这个行：

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

什么该行实际上指出的是，它基本上需要以下两个条件之一是True：

1. 如果 “1”（没有别的）中选择
2. 如果 “2”，而不是关键

这是一个典型的错误，如果你是一个初学者，如果你写的，你可以很容易地解决这个问题如下（最简单的解决）：

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

这意味着：如果选择是等于列表[1,2]中包含的任何元素，并且键不为真。

Answer 2

elif any(x in ["1", "2"] for x in choice) and not key: 

Answer 3

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

这interpretted如下（ “1” 或（（ “2” 中选择）和（非键）））

由于 “1” 总是为真，这总是如此。我想你的意思是：

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