python学习-游戏2

# coding:utf-8
#!/usr/bin/python
'''
Created on Jan 4, 2018

@author: wayne
'''
import random #导入模块
import time


def displayIntro():         #定义一个函数，不执行代码，在调用的时候才执行
		print('you are in a land full of dragons.In front of you')
		print('you see two caves.In one cave,the dragon is friendly')
		print('and will share his treasure with you the other dragon')
		print('is greedy and hungry,and will eat you on sight.')
		print()

def chooseCave():       #定义一个玩家选择要进入的1或者2的函数
		cave = ''           #创建一个新的空字符串
		while cave != '1' and cave != '2':
				'''
												布尔操作符，and，or，not
				and 表示：表达式全为true为true，其中一个值为False或者全为False 表达式结果为False
				>>> True and True
				True
				>>>True and False
				False
				>>> False and False
				False
				or:只要其中一个值为True 那么结果就为True ，只有两个值都为False的时候才是False
				>>>True or False
				True
				>>> True or True
				True
				>>>False or False
				False
				not :只能作用一个值，不能两个值组合在一起，not是取值的反值
				>>> not True
				False
				>>> not False
				True
				>>> not (1 ==2)
				True
				'''        
				print('which cave will you go into?(1 or 2)')
				cave = input() #接收输入值，
				return cave
def checkCave(chosenCave):  #继承chosenCave的函数值
		print('you approach the cave .....')
		time.sleep(2)
		print('it is dark and spooky.....')
		time.sleep(2)
		print('A large dragon jumps out in front of you!he opens his jaws and .....')
		print()
		time.sleep(2)

		friendlyCave = random.randint(1,2)  #随机产生一个数字，在1，2中间
		if chosenCave == str(friendlyCave): #判断输入值和随机产生的值是否一致，一致则玩家获胜，否则失败
				print('Gives you his treasure!')
		else:
				print('Gobbles you down in one bite!')
playAgain = 'yes'       #存储一个初始值，"yes"

while playAgain == 'yes' or playAgain == 'y': #判断条件是否为True
		displayIntro() #调用函数
		caveNumber = chooseCave()   #调用输入值的函数
		checkCave(caveNumber) #调用checkCave的函数
		print('Do you want to play again!(yes or no)') #判断玩家是否还要玩
		playAgain = input() #接收键盘输入的值，如果是True则循环开始