基于文本的冒险游戏库存问题python
基于文本的冒险游戏库存问题python
提问于 2013-11-03 23:47:00
我一直在做一个恐怖短信为基础的游戏&我有一些库存问题。清单是一个数组，我可以在任何函数中调用它。我到了那里，但每次都会用新的库存来重新填充数组。我需要一些帮助，这些是我的库存功能：
#Populating an aray with items to be used throughout the game.
def createItems():
items = range(0, 11)
if items[9] != "full":
items[1] = ("Axe")
items[2] = ("Gas")
items[3] = ("keys")
items[4] = ("gun")
items[5] = ("note")
items[9] = ("full")
return items
else:
return items
# this function is going to check if the item passed to it is still in the array
def checkItems(item):
list = createItems()
itemC = item
for i in range (0, 11):
if list[i] == itemC:
return ("no")
break
def createInventory():
inv = range(0 , 11)
inv[10] = ("made")
if inv[10] != ("made"):
for i in range (0, 11):
inv[i] = 0
return inv
def stockInventory(item):
inv = createInventory()
for i in range (0, 11):
if inv[i] == 0:
inv[i] = item
break
return inv
def checkInventory(item):
itemC = item
inv = createInventory()
for i in range(0, 11):
if itemC == inv[i]:
return ("yes")回答 2
Stack Overflow用户
回答已采纳
发布于 2013-11-04 00:30:47
这可能不是答案，但从代码和问题中我可以看出，这应该会有所帮助。注意您的代码中的差异&我的，并相应地进行更改。
# Main Inventory
Inventory = createInventory()
# Populating given inventory aray with items to be used throughout the game.
def createItems(inv):
items = inv
items[1] = "Axe"
items[2] = "Gas"
items[3] = "keys"
items[4] = "gun"
items[5] = "note"
items[9] = "full"
# Check if the item passed to it is still in the inventory array
def checkItems(item):
items = Inventory
for i in range(len(items)):
if items[i] == item:
return "yes"
return "no"
def createInventory():
inv = range(11)
inv[10] = "made"
return inv
def stockInventory(item):
inv = Inventory
for i in range (11):
if inv[i] == 0:
inv[i] = item
break
return inv
def checkInventory(item):
inv = Inventory
for i in range(0, 11):
if item == inv[i]:
return "yes"
return "no"Stack Overflow用户
发布于 2017-01-15 04:27:03
你可以把存货做成低端的价值。
inventory = {"sword":True, "axe":True}它也将有助于为您的库存制作课程。
class Weapon():
def attack(self, monster):
pass
class Axe(Weapon):
def __init__(self):
self.name = "axe"
self.damage = 1
def attack(self, monster):
monster.hp -= self.damage页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/19759646
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