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我想做一個簡單的遊戲。簡單的Python腳本中的流控制
邏輯是這樣的:「有五個門,每個門編號爲1到5,用戶將被要求輸入任意一個數字,例如,如果他們輸入」1「,GoldRoom將被打開(並且相關的類將被處理)。「
現在,我已經定義了一個類GoldRoom()
,並進行了測試,輸入了「1」。處理過程按預期進行。但是,當我輸入「2」作爲我的選擇時,處理仍然發生,而不是print語句,即else語句沒有執行。
我哪裏錯了?
#################################
# Learning to make a game#
#################################
# An attempt to make a game
# Each room will be described by a class, whose base class will be Room
# The user will be prompted to enter a number, each number will be assigned with a Room in return
from sys import exit
print "Enter your choice:"
room_chosen = int(raw_input("> "))
if room_chosen == 1:
goldroom = GoldRoom()
goldroom.gold_room()
def dead(why):
print "why, Good Job!"
exit(0)
#class Room(object): #the other room will be derived of this
# pass
class Room(object):
pass
class GoldRoom(Room):
# here the user will be asked with question on how much Gold he wants
print"This room is full of gold. How much do you take!"
next = raw_input("> ")
if "0" in next or "1" in next:
how_much = int(next)
print how_much
else:
dead("Man, learn to type some number")
if how_much < 50:
print "Nice, you are not greedy, you win!"
exit(0)
else:
dead("You greedy bastard!")
#class KoiPondRoom(Room):
# in this room, the user will be made to relax
#class Cthulhu_Room(Room):
# sort of puzzle to get out
#class Bear_Room(Room):
# bear room
#class Dark_Room(Room):
# Dark Room, will be turned into Zombie
#class Dead_Room(Room):
# Those who enter here would be dead
if room_chosen == 1:
goldroom = GoldRoom()
goldroom.gold_room()
else:
print "YOU SUCK!"
哦,好的,謝謝你的幫助:) ..我錯過了那個班,但初始化它:) – peterparker
但現在我遇到了一個錯誤: – peterparker
PS C:\ USers \ rk \ lpthw> python ex45.py在 類戈爾德魯姆(室) 文件 「ex45.py」 第24行:輸入您的選擇: > 1 回溯(最近通話最後一個) 文件 「ex45.py」 32行,在戈爾德魯姆 如果在接下來的 「0」 或在下一個 「1」: 類型錯誤:類型 'builtin_function_or_method' 的參數不是可迭代 –
peterparker