blocks|key|773552|text|当用户按下中断键Ctrl-C时，会引发KeyboardInterrupt异常。在python中，这是从SIGINT信号转换而来的。这意味着，您可以使用信号模块随心所欲地处理它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|773553|import+signal

def+handler(signum,+frame):
++++print("do+whatever,+like+call+thread.interrupt_main()")

signal.signal(signal.SIGINT,+handler)
print("Waiting+for+SIGINT...")
signal.pause()|code-block|syntax|javascript|773554|这样，您就可以在收到键盘中断时做任何您想做的事情。|773555|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The KeyboardInterrupt exception is raised when a user hits the interrupt key, Ctrl-C. In python this is translated from a SIGINT signal. That means, you can get handle it however you want using the signal module:
<pre><code>import signal

def handler(signum, frame):
 print(&quot;do whatever, like call thread.interrupt_main()&quot;)

signal.signal(signal.SIGINT, handler)
print(&quot;Waiting for SIGINT...&quot;)
signal.pause()
</code></pre>
That way, you can do whatever you want at the receipt of a keyboard interrupt.

blocks|key|988891|text|我在Linux下用python2.5、2.6、3版本测试了你的代码，当按下CTRL-C时，所有代码都抛出了"KeyboardInterrupt“异常。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|988892|也许某个异常处理捕捉到了中断，或者您的问题是这样的：Why+is+KeyboardInterrupt+not+working+in+python?|offset|length|988893|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/543534/why-keyboardinterrupt-is-not-working-in-python^0|0|Q|1B|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

I tried your code with python versions 2.5, 2.6, 3 under Linux and all throw "KeyboardInterrupt" exception when hitting CTRL-C.

Maybe some exception handling catches the Interrupt or your problem is like this:
<a href="https://stackoverflow.com/questions/543534/why-keyboardinterrupt-is-not-working-in-python">Why is KeyboardInterrupt not working in python?</a>

blocks|key|773641|text|最优雅的解决方案当然是threading.Event，不过如果你只需要快速修改一下，下面的代码就能很好地工作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|773642|import+time

def+main():
++++print("It’s+time !")

if+__name__+==+"__main__":
++++print("press+ctrl-c+to+stop")
++++loop_forever+=+True
++++while+loop_forever:
++++++++main()
++++++++try:
++++++++++++time.sleep(60)
++++++++except+KeyboardInterrupt:
++++++++++++loop_forever+=+False|code-block|syntax|javascript|773643|entityMap^0|B|F|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The most elegant solution is certainly <code>threading.Event</code>, though if you only need a quick hack, this code works well :

<pre><code>import time

def main():
 print("It’s time !")

if __name__ == "__main__":
 print("press ctrl-c to stop")
 loop_forever = True
 while loop_forever:
 main()
 try:
 time.sleep(60)
 except KeyboardInterrupt:
 loop_forever = False
</code></pre>

blocks|key|988980|text|我想我应该把这个扔进去。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|988981|import+time


def+sleep(seconds):
++++for+i+in+range(seconds):
++++++++try:
++++++++++++time.sleep(1)
++++++++except+KeyboardInterrupt:
++++++++++++print("Oh!+You+have+sent+a+Keyboard+Interrupt+to+me.\nBye,+Bye")
++++++++++++break


sleep(60)|code-block|syntax|javascript|988982|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Figured I'd throw this in.
<pre><code>import time


def sleep(seconds):
 for i in range(seconds):
 try:
 time.sleep(1)
 except KeyboardInterrupt:
 print(&quot;Oh! You have sent a Keyboard Interrupt to me.\nBye, Bye&quot;)
 break


sleep(60)
</code></pre>

blocks|key|989010|text|基于@Andreas+Jung的回答|type|unstyled|depth|inlineStyleRanges|entityRanges|data|989011|++++++++for+i+in+range(360):
++++++++++++try:
++++++++++++++++sleep(1)
++++++++++++except+KeyboardInterrupt:
++++++++++++++++sys.exit(0)|code-block|syntax|javascript|989012|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Based on @Andreas Jung answer
<pre><code> for i in range(360):
 try:
 sleep(1)
 except KeyboardInterrupt:
 sys.exit(0)
</code></pre>

I need to break from time.sleep() using ctrl c.

<pre><code>While 1:
 time.sleep(60)
</code></pre>

In the above code when the control enters time.sleep function an entire 60 seconds needs to elapsed for python to handled the CTRL C 

Is there any elegant way to do it. such that I can interrupt even when the control is in time.sleep function

edit

I was testing it on a legacy implementation which uses python 2.2 on windows 2000 which caused all the trouble . If I had used a higher version of python CTRL C would have interrupted the sleep() . I did a quick hack by calling sleep(1) inside a for loop . which temporarily fixed my issue

break/interrupt a time.sleep() in python

Linux

我需要使用ctrl c来中断time.sleep()。While 1:    time.sleep(60)在上面的代码中，当控件进入time.sleep函数时，python需要整整60秒来处理CTRL C有什么优雅的方法可以做到这一点吗？这样，即使控件处于time.sleep函数中，我也可以中断编辑我是在一个遗留的实现...

问在python中中断/中断time.sleep()
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在python中中断/中断time.sleep()EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在python中中断/中断time.sleep()
EN