标签:python python-2-7 traceback exit
为学习目的构建一个简单的“Rock,Paper,Scissors”Python游戏.
我已经阅读了一些关于退出Python而没有回溯的其他帖子.我正在尝试实现它,但仍然得到追溯!有些Python wiz可以指出这个Python假人有什么不对吗?想法是单击RETURN(或键入“yes”或“y”将使程序再次运行play(),但是按CTRL-C将关闭它而没有回溯.我使用的是Python 2.7.
# modules
import sys, traceback
from random import choice
#set up our lists
ROCK, PAPER, SCISSORS = 1, 2, 3
names = 'ROCK', 'PAPER', 'SCISSORS'
#Define a function for who beats who?
def beats(a, b):
return (a,b) in ((PAPER, ROCK), (SCISSORS, PAPER), (ROCK, SCISSORS))
def play():
print "Please select: "
print "1 Rock"
print "2 Paper"
print "3 Scissors"
# player choose Rock, Paper or Scissors
player_choice = int(input ("Choose from 1-3: "))
# assigns the CPU variable a random CHOICE from a list.
cpu_choice = choice((ROCK, PAPER, SCISSORS))
if cpu_choice != player_choice:
if beats(player_choice, cpu_choice):
print "You chose %r, and the CPU chose %r." % (names[player_choice - 1], names[cpu_choice - 1])
print "You win, yay!!"
else:
print "You chose %r, and the CPU chose %r." % (names[player_choice - 1], names[cpu_choice - 1])
print "You lose. Yuck!"
else:
print "You chose %r, and the CPU chose %r." % (names[player_choice - 1], names[cpu_choice - 1])
print "It's a tie!"
print "Do you want to play again? Click RETURN to play again, or CTRL-C to exit!"
next = raw_input("> ")
# THIS IS WHAT I'M WORKING ON - NEED TO REMOVE TRACEBACK!
if next == "yes" or "y":
try:
play()
except KeyboardInterrupt:
print "Goodbye!"
except Exception:
traceback.print_exc(file=sys.stdout)
sys.exit(0)
elif next == None:
play()
else:
sys.exit(0)
# initiate play() !
play()
解决方法:
你调用play()两次,所以你需要将两个case放在try / except块中:
if next in ("yes", "y"):
try:
play()
except KeyboardInterrupt:
print "Goodbye!"
except Exception:
traceback.print_exc(file=sys.stdout)
sys.exit(0)
elif next is None:
try:
play()
except KeyboardInterrupt:
print "Goodbye!"
except Exception:
traceback.print_exc(file=sys.stdout)
sys.exit(0)
else:
sys.exit(0)
我已经纠正了另外两个问题,最好在python中测试None,而你的第一个if测试不起作用,因为next ==“yes”或“y”被解释为next ==“yes “与”y“分开,与之间或之间. “y”始终被视为True,因此您永远不会来到代码中的其他分支.
请注意,我怀疑上面的代码可以简化得更多,但你根本没有告诉我们你的play()函数,所以你让我们猜测你想要做什么.
标签:python,python-2-7,traceback,exit 来源: https://codeday.me/bug/20190723/1511787.html
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