基于Python文本的冒险游戏,需要帮助

Question

我可以用一些帮助搞清楚为什么我的游戏不允许我拿起一把剑.出现的错误消息是

回溯(最近一次调用最后一次):文件"",第36行,在文件"",第1行,在NameError中:名称'y'未定义

这是基本游戏,我对此非常新,所以简单的术语会很好.

print 'You enter a Dungeon with three doors. Do you want to enter door #1 door #2 or door #3?'
door = raw_input('> ')

if door == "1":
    print 'Theres a dragon eating a human, the dragon is massive and looks terrifying'
    print 'what do you want to do?'
    print '#1 try to save the human'
    print '#2 scream and run around'
    dragon = raw_input('> ')

    if dragon == "1":
        print 'You approach the dragon sneakily. After what feel like a thousand years you are finally close to the dragon. You try to step closer. The dragon shifts and crushes you. Well done!'

    elif dragon == "2":
        print 'You scream and run around like a chicken with no head. The dragon is not impressed with you. It snorts and engulfs you in flames. You die a painful and worthless death. Well done!'

elif door == "2":
    print 'You stare into a deep dark cave.'
    print 'Oh dear, it seems you have been driven you quite insane.'
    print '#1. drool'
    print '#2. scream, drool, and walk forward'
    print '#3. Understand computer programming completely and get an A plus'
    insanity = raw_input('> ')

    if insanity == "1":
        print 'Your body survives but your mind does not, you drool for eternity!'

    if insanity == "2":
        print 'You slip on a mysterious liquid, then fall into a deep sleep, you awake years later at the kiss of a prince. Yay!'

    if insanity == "3":
        print 'Congradulations! You passed Programming! Time to graduate and face the real world.. So I guess you actully lost. Sorry.'

elif door == "3":
    print ("You emerge into a brighter area and see a small sword lying on the ground")
    ch1 = str(input("Do you take it? [y/n]:"))
    if ch1 == "y":
        print 'You have taken the sword!'
        stick = 1

else:
    print 'You have chosen not to take the sword'
    sword = 0

Answer 1

ch1 = str(input("Do you take it? [y/n]:"))

如果您使用input并且用户键入"y",则解释器将搜索变量y并返回其值.你没有y变量,所以它崩溃了.你可能想在raw_input这里使用,而不是input.

ch1 = str(raw_input("Do you take it? [y/n]:"))

顺便说一句,raw_input无论如何返回一个字符串,所以你真的不需要这个str电话.

ch1 = raw_input("Do you take it? [y/n]:")