Python 在扭曲程序中中断原始输入
我将提及这一点和这一点: 所以我现在做的是:Python 在扭曲程序中中断原始输入,python,twisted,interrupt,keyboard-events,keyboardinterrupt,Python,Twisted,Interrupt,Keyboard Events,Keyboardinterrupt,我将提及这一点和这一点: 所以我现在做的是: def interrupted(signum, stackframe): log.warning('interrupted > Got signal: %s', signum) menu.quitMenu = True # to stop my code signal.signal(signal.SIGINT, interrupted) # Handle KeyboardInterrupt 问题是,虽然菜单被通知必须停止,并
def interrupted(signum, stackframe):
log.warning('interrupted > Got signal: %s', signum)
menu.quitMenu = True # to stop my code
signal.signal(signal.SIGINT, interrupted) # Handle KeyboardInterrupt
问题是,虽然菜单被通知必须停止,并且很快就会停止,但它现在不能这样做,因为它在原始输入中被阻塞了:
def askUser(self):
current_date = datetime.now().isoformat(' ')
choice = raw_input('%s > ' % current_date)
return choice
因此,由于twisted正在删除默认的中断处理程序,raw_input
不会停止。我仍然需要在^C
之后按enter
,它才能停止
我如何强制停止原始输入,而不安装默认的中断处理程序,这是twisted上下文中问题的根源(因为twisted本身不希望被中断)
我想问题不只是与原始输入有关:任何占用无限时间(或超过预设限制)的函数都应该以某种方式中断
是否有一个可接受的扭曲模式
编辑
这是完整的测试代码:
from datetime import datetime
class Menu:
def __init__(self):
self.quitMenu = False
def showMenu(self):
print '''
A) Do A
B) Do B
'''
def askUser(self):
current_date = datetime.now().isoformat(' ')
choice = raw_input('%s > Please select option > ' % current_date)
print
return choice
def stopMe(self):
self.quitMenu = True
def alive(self):
return self.quitMenu == False
def doMenuOnce(self):
self.showMenu()
choice = self.askUser()
if not self.alive() : # Maybe somebody has tried to stop the menu while in askUser
return
if choice == 'A' : print 'A selected'
elif choice == 'B' : print 'B selected'
else : print 'ERR: choice %s not supported' % (choice)
def forever(self):
while self.alive():
self.doMenuOnce()
from twisted.internet import reactor, threads
import signal
class MenuTwisted:
def __init__(self, menu):
self.menu = menu
signal.signal(signal.SIGINT, self.interrupted) # Handle KeyboardInterrupt
def interrupted(self, signum, stackframe):
print 'Interrupted!'
self.menu.stopMe()
def doMenuOnce(self):
threads.deferToThread(self.menu.doMenuOnce).addCallback(self.forever)
def forever(self, res=None):
if self.menu.alive() :
reactor.callLater(0, self.doMenuOnce)
else :
reactor.callFromThread(reactor.stop)
def run(self):
self.forever()
reactor.run()
我可以用两种不同的方式运行
正常方式:
按^C
立即停止程序:
A) Do A
B) Do B
2013-12-03 11:00:26.288846 > Please select option > ^CTraceback (most recent call last):
File "twisted_keyboard_interrupt.py", line 72, in <module>
menu.forever()
File "twisted_keyboard_interrupt.py", line 43, in forever
self.doMenuOnce()
File "twisted_keyboard_interrupt.py", line 34, in doMenuOnce
choice = self.askUser()
File "twisted_keyboard_interrupt.py", line 22, in askUser
choice = raw_input('%s > Please select option > ' % current_date)
KeyboardInterrupt
按^C
将产生:
A) Do A
B) Do B
2013-12-03 11:04:18.678219 > Please select option > ^CInterrupted!
但是askUser
实际上并没有被中断:我仍然需要按enter
来完成raw\u input
。正确的处理方法是,而不是试图使阻塞输入功能可中断。换句话说,raw_input
从根本上说是您所攻击的问题的错误解决方案
然而,如果你真的只是想了解这里发生了什么,诀窍是在调用reactor.callFromThread(reactor.stop)
之后,你必须以某种方式提示原始输入退出;这不是正常的。然而,由于您在线程中运行它,它实际上根本不可中断,因为在Python中。所以我认为你想要的可能实际上是不可能的。我相信关闭sys.stdin
可能会从raw\u input
下拉出地毯,即使它在一个线程中,但底层库似乎做了比简单地从FD读取更聪明的事情,因此关闭它没有什么好处
menu = Menu()
menutw = MenuTwisted(menu)
menutw.run()
A) Do A
B) Do B
2013-12-03 11:04:18.678219 > Please select option > ^CInterrupted!