python – Ctrl-C结束我的脚本,但它没有被KeyboardInterrupt异常捕获

印度阿三17 2019-07-03

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我有一个python脚本,包含一个大循环读取文件和做一些东西(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,当我按下Ctrl-C时程序停止但是它没有被我的两个KeyboardInterrupt异常中的任何一个捕获,虽然我确定它当前在循环中(因此至少在big try / except中).

怎么可能？起初我以为是因为我正在使用的其中一个包没有正确处理异常(例如只使用“except：”),但如果是这样,我的脚本就不会停止.但脚本会停止,它应该被至少一个我的两个捕获,除了,对吧？

我哪里错了？

提前致谢！

编辑：

通过在try-except之后添加finally：子句并在两个try-except块中打印回溯,当我按下Ctrl-C时它通常显示None,但我曾经设法得到它(似乎它来自urllib2,但我不知道是不是我无法捕获KeyboardInterrupt的原因：

Traceback(最近一次调用最后一次)：

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>

解决方法:

我已经在我的评论中提出了这个问题,这个问题可能是由问题中遗漏的代码部分引起的.但是,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.

您在评论中提到使用了samppipe Python包.这个Python包使用JPype来创建绑定到Java的语言…我可以使用以下Python程序重现您的问题：

from boilerpipe.extract import Extractor
import time

try:
  for i in range(10):
    time.sleep(1)

except KeyboardInterrupt:
  print "Keyboard Interrupt Exception"

如果使用Ctrl-C中断此程序,则不会抛出异常.似乎程序立即终止,而Python解释器没有机会抛出异常.当删除导管的导入时,问题就消失了……

与gdb的调试会话表明,如果导入了boilerpipe,Python会启动大量线程：

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samppipe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

因此,我假设您的Ctrl-C信号在另一个线程中处理,或者jpype执行其他破坏Ctrl-C处理的奇怪事情.

编辑：作为一种可能的解决方法,您可以注册一个信号处理程序,捕获当您按Ctrl-C时进程收到的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,您将收到通知,并且您将能够在程序的中心点处理该事件.如果要在此处理程序中,可以终止脚本.如果不这样做,脚本将在信号处理函数返回后继续运行.请参阅以下示例：

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."
#    extractor = Extractor(extractor='ArticleExtractor', url=your_url)
except KeyboardInterrupt:
    print "Keyboard Interrupt Exception"

来源：https://www./content-1-293401.html