简要咨询咨询QQ网站导航网站搜索手机站点联系我们设为首页加入收藏 

python实现的简单文本类游戏实现方法

来源:易贤网   阅读:1354 次  日期:2015-04-30 14:21:33

温馨提示:易贤网小编为您整理了“python实现的简单文本类游戏实现方法”,方便广大网友查阅!

本文实例讲述了python实现的简单文本类游戏实现方法。分享给大家供大家参考。具体实现方法如下:

############################################################

# - My version on the game "Dragon Realm".

# - taken from the book "invent with python" by Al Sweigart.

# - thanks for a great book Mr Sweigart.

# - this code takes advantage of python 3.

############################################################

#files.py

import random

import time

print('\n\n[--system--] one file is bad the other is good ..guess the right one.\n')

print('\n\nconnecting....')

time.sleep(1)

print('....')

time.sleep(1)

print('....')

time.sleep(1)

print('....')

time.sleep(1)

print('\nconnection established')

def displayIntro():

print('------------')

print('SYSTEM FILES')

print('------------\n')

print('1.) file.')

print('2.) file.\n')

def chooseOption():

option = ''

while option != '1' and option != '2':

print('which file to download? 1 or 2')

option = input('user:> ')

return option

def checkOption(chosenOption):

print('\nintialising download....')

time.sleep(1)

print('accessing file....')

time.sleep(1)

print('downloading....')

time.sleep(1)

print('....')

time.sleep(1)

print('....')

time.sleep(1)

goodfile = random.randint(1, 2)

if chosenOption == str(goodfile):

print('\ndownload complete.')

print('\nGAME OVER')

else:

print('\nfile corrupt')

print('system infected.')

print('\nGAME OVER')

playAgain = 'yes'

while playAgain == 'yes':

displayIntro()

optionNumber = chooseOption()

checkOption(optionNumber)

print('\ndownload again? .... (yes or no)')

playAgain = input('user:> ')

############################################################

# - My version of the game "guess the number".

# - taken from the book "invent with python" by Al Sweigart.

# - thanks for a great book Mr Sweigart.

# - this code takes advantage of python 3.

############################################################

# -NOTE - this program will crash if a number is not typed.

#digitcode.py

import random

import time

guessesTaken = 0

print('\n\n\n\n\n[--system--] enter code in 15 trys to avoid lockout\n')

print('\nconnecting....')

time.sleep(1)

print('....')

time.sleep(1)

print('....')

time.sleep(1)

print('....')

time.sleep(1)

print('connection established\n')

print('---------------------')

print(' MAINFRAME - LOGIN ')

print('---------------------')

print('\nenter 3 digit access code..')

number = random.randint(000, 999)

while guessesTaken < 15:

print()

guess = input('user:> ')

guess = int(guess)

guessesTaken = guessesTaken + 1

if guess < number:

print('\nACCESS - DENIED -code to low')

if guess > number:

print('\nACCESS - DENIED -code to high')

if guess == number:

break

if guess == number:

guessesTaken = str(guessesTaken)

print('\nverifying ....')

time.sleep(1)

print('\nauthenticating ....')

time.sleep(1)

print('....')

time.sleep(1)

print('....')

time.sleep(1)

print('\nACCESS - GRANTED')

print('\nGAME OVER\n')

exit(0)

if guess != number:

number = str(number)

print('\n....')

time.sleep(1)

print('\n....')

time.sleep(1)

print('\nSYSTEM LOCKED -the code was ' + number)

print()

exit(0)

希望本文所述对大家的Python程序设计有所帮助。

更多信息请查看IT技术专栏

更多信息请查看网络编程
点此处就本文及相关问题在本站进行非正式的简要咨询(便捷快速)】     【点此处查询各地各类考试咨询QQ号码及交流群
上一篇:解析Python下的多进程编程
下一篇:python使用append合并两个数组的方法
易贤网手机网站地址:python实现的简单文本类游戏实现方法
由于各方面情况的不断调整与变化,易贤网提供的所有考试信息和咨询回复仅供参考,敬请考生以权威部门公布的正式信息和咨询为准!