python实现的简单文本类游戏实例-파이썬 튜토리얼-php.cn

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python实现的简单文本类游戏实例

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Jun 10, 2016 pm 03:14 PM

python

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&#63; 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&#63; .... (yes or no)')
  playAgain = input('user:> ')

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############################################################
# - 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)

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希望本文所述对大家的Python程序设计有所帮助。

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