日韩黑丝制服一区视频播放|日韩欧美人妻丝袜视频在线观看|九九影院一级蜜桃|亚洲中文在线导航|青草草视频在线观看|婷婷五月色伊人网站|日本一区二区在线|国产AV一二三四区毛片|正在播放久草视频|亚洲色图精品一区

分享

python – Ctrl-C結(jié)束我的腳本,但它沒有被KeyboardInterrupt異常捕獲

 印度阿三17 2019-07-03

我有一個python腳本,包含一個大循環(huán)讀取文件和做一些東西(我使用幾個包,如urllib2,httplib2或BeautifulSoup).

它看起來像這樣:

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

問題是,當我按下Ctrl-C時程序停止但是它沒有被我的兩個KeyboardInterrupt異常中的任何一個捕獲,雖然我確定它當前在循環(huán)中(因此至少在big try / except中).

怎么可能?起初我以為是因為我正在使用的其中一個包沒有正確處理異常(例如只使用“except:”),但如果是這樣,我的腳本就不會停止.但腳本會停止,它應該被至少一個我的兩個捕獲,除了,對吧?

我哪里錯了?

提前致謝!

編輯:

通過在try-except之后添加finally:子句并在兩個try-except塊中打印回溯,當我按下Ctrl-C時它通常顯示None,但我曾經(jīng)設法得到它(似乎它來自urllib2,但我不知道是不是我無法捕獲KeyboardInterrupt的原因:

Traceback(最近一次調(diào)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>

解決方法:

我已經(jīng)在我的評論中提出了這個問題,這個問題可能是由問題中遺漏的代碼部分引起的.但是,確切的代碼不應該是相關(guān)的,因為當Python代碼被Ctrl-C中斷時,Python通常會拋出KeyboardInterrupt異常.

您在評論中提到使用了samppipe Python包.這個Python包使用JPype來創(chuàng)建綁定到Java的語言…我可以使用以下Python程序重現(xiàn)您的問題:

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中斷此程序,則不會拋出異常.似乎程序立即終止,而Python解釋器沒有機會拋出異常.當刪除導管的導入時,問題就消失了……

與gdb的調(diào)試會話表明,如果導入了boilerpipe,Python會啟動大量線程:

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)]

沒有samppipe導入的gdb會話:

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,我假設您的Ctrl-C信號在另一個線程中處理,或者jpype執(zhí)行其他破壞Ctrl-C處理的奇怪事情.

編輯:作為一種可能的解決方法,您可以注冊一個信號處理程序,捕獲當您按Ctrl-C時進程收到的SIGINT信號.即使導入了boilerpipe和JPype,信號處理程序也會被觸發(fā).這樣,當用戶按下Ctrl-C時,您將收到通知,并且您將能夠在程序的中心點處理該事件.如果要在此處理程序中,可以終止腳本.如果不這樣做,腳本將在信號處理函數(shù)返回后繼續(xù)運行.請參閱以下示例:

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" 
來源:https://www./content-1-293401.html

    本站是提供個人知識管理的網(wǎng)絡存儲空間,所有內(nèi)容均由用戶發(fā)布,不代表本站觀點。請注意甄別內(nèi)容中的聯(lián)系方式、誘導購買等信息,謹防詐騙。如發(fā)現(xiàn)有害或侵權(quán)內(nèi)容,請點擊一鍵舉報。
    轉(zhuǎn)藏 分享 獻花(0

    0條評論

    發(fā)表

    請遵守用戶 評論公約

    類似文章 更多