Python 中的文本游戏问题。循环输入提示

Question

我的下一个问题是我的一个学生在他的基于文本的游戏中的代码。这个想法是，在游戏中你可以拿起一个物品并将它放在一个选定的口袋里。问题是，当他运行这段代码时，他得到了选择一个口袋来放入物品的提示，但随后输入提示只是循环出现关于将物品放在哪里的相同问题。

想法？

class grab():
    #Checks the player's pockets for items.
    def pocket_check(inventory, pocket, thing, input, state, list, room):
        if inventory[pocket] != None:
            print("You find that your %s is stuffed full.")%(input)
        if inventory[pocket] == None:
            inventory[pocket] = thing
            del room["Items"][thing]
            state["Strain"] += list[thing]["Weight"]
            print("You put the %s in your %s.")%(thing, input)

    #Takes items and putting them in the player's inventory
    def grab(inventory, thing, list, state, room):
        go = True
        while go:
            if list != "WEAPON" and state["Strain"]+list[thing]["Weight"] <= state["Strength"]:
                inp = input("Where will you put it?").lower()
                inp_broke = inp.split(" ")
                if inp_broke[0] == "stop":
                    go = False
                elif inp_broke[0] != "put":
                    inp = input("Where will you PUT it?").lower()
                    inp_broke = inp.split(" ")
                elif inp_broke[0] == "put":
                    if inp_broke[1:3] == "shirt pocket":
                        pocket_check(inventory, "Shirt Pocket", thing, "shirt pocket", state, list, room)
                    if inp_broke[1:4] == "left front pocket":
                        pocket_check(inventory, "Left Front Pocket", thing, "left front pocket", state, "ITEM", room)
                    if inp_broke[1:4] == "right front pocket":
                        pocket_check(inventory, "Right Front Pocket", thing, "right front pocket", state, "ITEM", room)
                    if inp_broke[1:3] == "back pocket":
                        pocket_check(inventory, "Back Pocket", thing, "back pocket", state, "ITEM", room)

Answer 1

来自您的代码段：

np_broke = inp.split(" ")
...
if inp_broke[1:3] == "shirt pocket":

从 split() 分割列表 returns 字符串列表，而不是字符串：

>>> inp = "put shirt pocket"
>>> inp_broke = inp.split(" ")
>>> inp_broke[1:3]
['shirt', 'pocket']

Answer 2

这个

if inp_broke[1:3] == "shirt pocket":

行不通。 inp_broke 是一个（字符串的）列表，并不会神奇地组合成一个字符串。您可以使用 str.join 方法：

if " ".join(inp_broke[1:3]) == "shirt pocket":

这会将列表的各个项目组合成一个字符串，由初始字符串分隔（在本例中为单个 space）。

Answer 3

while go 将永远循环，除非在循环中将 False 分配给 go。 pocket_check 方法可以 return 一个布尔值，这取决于它的成功和调用时分配给 go 的 return 值。

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