Ctrl-C结束了我的脚本，但它没有被KeyboardInterrupt异常捕获

Question

我有一个Python脚本，其中包含一个大循环读取文件并做一些事情(我使用几个包，如urllib2, httplib2或BeautifulSoup)。

它看起来像这样:

try:
    with open(fileName, 'r') as file :
        for i, line in enumerate(file):
            try:
                # a lot of code
                # ....
                # ....
            except urllib2.HTTPError:
                print "n >>> HTTPError"
            # a lot of other exceptions
            # ....
            except (KeyboardInterrupt, SystemExit):
                print "Process manually stopped"
                raise
            except Exception, e:
                print(repr(e))
except (KeyboardInterrupt, SystemExit):
    print "Process manually stopped"
    # some stuff

问题是，当我按Ctrl+C时，程序停止，但它没有被我的两个KeyboardInterrupt异常捕获，尽管我确信它目前在循环中(因此至少在大try/except中)。

这怎么可能?起初，我认为这是因为我使用的一个包没有正确处理异常(比如通过使用"except:"only)，但如果是这样的话，我的脚本不会停止。但是脚本确实停止了，并且它应该被至少一个(我的两个除外)捕获，对吗?

我错在哪里?

提前感谢!

编辑:

在try-except之后添加finally:子句并在两个try-except块中打印跟踪，当我点击Ctrl+C时，它通常显示None，但我曾经设法得到这个(似乎它来自urllib2，但我不知道这是否是我无法捕获KeyboardInterrupt的原因):

回溯(最近一次调用):

File "/home/darcot/code/Crawler/crawler.py", line 294, in get_articles_from_file
  content = Extractor(extractor='ArticleExtractor', url=url).getText()
File "/usr/local/lib/python2.7/site-packages/boilerpipe/extract/__init__.py", line 36, in __init__
  connection  = urllib2.urlopen(request)
File "/usr/local/lib/python2.7/urllib2.py", line 126, in urlopen
  return _opener.open(url, data, timeout)
File "/usr/local/lib/python2.7/urllib2.py", line 391, in open
  response = self._open(req, data)
File "/usr/local/lib/python2.7/urllib2.py", line 409, in _open
  '_open', req)
File "/usr/local/lib/python2.7/urllib2.py", line 369, in _call_chain
  result = func(*args)
File "/usr/local/lib/python2.7/urllib2.py", line 1173, in http_open
  return self.do_open(httplib.HTTPConnection, req)
File "/usr/local/lib/python2.7/urllib2.py", line 1148, in do_open
  raise URLError(err)
URLError: <urlopen error [Errno 4] Interrupted system call>

Answer 1

我已经在对问题的评论中建议，这个问题很可能是由问题中遗漏的代码部分引起的。然而，确切的代码不应该是相关的，因为当Python代码被Ctrl-C中断时，Python通常会抛出KeyboardInterrupt异常。

您在评论中提到您使用boilerpipe Python包。这个Python包使用JPype创建语言绑定到Java…我可以用下面的Python程序重现你的问题:

from boilerpipe.extract import Extractor
import time
try:
  for i in range(10):
    time.sleep(1)
except KeyboardInterrupt:
  print "Keyboard Interrupt Exception"

如果用Ctrl-C中断程序，则不会抛出异常。似乎程序立即终止，Python解释器没有机会抛出异常。当删除boilerpipe的导入后，问题就消失了…

使用gdb的调试会话表明，如果导入boilerpipe, Python将启动大量线程:

gdb --args python boilerpipe_test.py
[...]
(gdb) run
Starting program: /home/fabian/Experimente/pykeyinterrupt/bin/python boilerpipe_test.py
warning: Could not load shared library symbols for linux-vdso.so.1.
Do you need "set solib-search-path" or "set sysroot"?
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/usr/lib/libthread_db.so.1".
[New Thread 0x7fffef62b700 (LWP 3840)]
[New Thread 0x7fffef52a700 (LWP 3841)]
[New Thread 0x7fffef429700 (LWP 3842)]
[New Thread 0x7fffef328700 (LWP 3843)]
[New Thread 0x7fffed99a700 (LWP 3844)]
[New Thread 0x7fffed899700 (LWP 3845)]
[New Thread 0x7fffed798700 (LWP 3846)]
[New Thread 0x7fffed697700 (LWP 3847)]
[New Thread 0x7fffed596700 (LWP 3848)]
[New Thread 0x7fffed495700 (LWP 3849)]
[New Thread 0x7fffed394700 (LWP 3850)]
[New Thread 0x7fffed293700 (LWP 3851)]
[New Thread 0x7fffed192700 (LWP 3852)]

没有boilerpipe导入的gdb会话:

gdb --args python boilerpipe_test.py
[...]
(gdb) r
Starting program: /home/fabian/Experimente/pykeyinterrupt/bin/python boilerpipe_test.py
warning: Could not load shared library symbols for linux-vdso.so.1.
Do you need "set solib-search-path" or "set sysroot"?
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/usr/lib/libthread_db.so.1".
^C
Program received signal SIGINT, Interrupt.
0x00007ffff7529533 in __select_nocancel () from /usr/lib/libc.so.6
(gdb) signal 2
Continuing with signal SIGINT.
Keyboard Interrupt Exception
[Inferior 1 (process 3904) exited normally 

所以我假设你的Ctrl-C信号是在一个不同的线程中处理的，或者jpype做了其他奇怪的事情，破坏了Ctrl-C的处理。

EDIT:作为一种可能的解决方案，您可以注册一个信号处理程序，该处理程序捕获进程在按Ctrl-C时接收的SIGINT信号。即使导入了boilerpipe和JPype，信号处理程序也会被触发。这样，当用户按下Ctrl-C时，您将得到通知，并且您将能够在程序的中心点处理该事件。如果需要，可以在此处理程序中终止脚本。如果不这样做，一旦信号处理程序函数返回，脚本将继续在中断的地方运行。请看下面的例子:

from boilerpipe.extract import Extractor
import time
import signal
import sys
def interuppt_handler(signum, frame):
    print "Signal handler!!!"
    sys.exit(-2) #Terminate process here as catching the signal removes the close process behaviour of Ctrl-C
signal.signal(signal.SIGINT, interuppt_handler)
try:
    for i in range(10):
        time.sleep(1)
#    your_url = "http://www.zeit.de"
#    extractor = Extractor(extractor='ArticleExtractor', url=your_url)
except KeyboardInterrupt:
    print "Keyboard Interrupt Exception" 

Answer 2

最有可能的是，当您的脚本在try块之外时，您发出了CTRL-C，因此没有捕捉到信号。