Question

我有一个python脚本，其中包含一个读取文件并执行某些操作的大循环（我使用的是几个包，如urllib2，httplib2或BeautifulSoup）。

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

问题是程序在我按下Ctrl-C时停止但是它没有被我的两个KeyboardInterrupt异常中的任何一个抓住，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使用的其中一个软件包没有正确处理异常（比如使用＆＃34;除了：＆＃34;只有）但是如果是这样的话，我的剧本不会停止。但脚本会停止，它至少应该被我的两个人抓住，除了，对吗？

我哪里错了？

提前致谢！

修改

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时它通常会显示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）：

追踪（最近一次呼叫最后一次）：

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>

Answer 1

我已在评论中提出这个问题，这个问题可能是由问题中遗漏的代码部分引起的。但是，确切的代码不应该相关，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。删除导入boilerpipe后，问题就会消失......

使用gdb的调试会话表示，如果导入boilerpipe，则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gdb导入的

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

Answer 2

最有可能当你的脚本在try块之外时你发出了CTRL-C，因此没有捕获信号。

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

2 个答案: