Question

我正在使用“灰色请求”模块,以消耗“Twitter”下流的“APIC”号。

self.conn = self.session.post(url, data=self.parameters, headers=self.headers)
    print >> sys.stderr,  Connected to Twitter Streaming API 
    try:
        for line in self.conn.iter_content(self.ITER_BYTES):
            self.status += line
            if line.endswith(self.DIVIDER) and self.status.strip():
                self.handler.handle(self.status)
                self.status = ""
    except Exception as e:
        pass

当我用一个键盘Interrupt或通过它发出终止信号时,我正经历以下四点错误:

^CTraceback (most recent call last):
  File "python-test.py", line 18, in <module>
    api.start()
  File "bin/polygraph/api/twitter/streaming.py", line 95, in start
    self.conn = self.session.post(url, data=self.parameters, headers=self.headers)
  File "venv/lib/python2.7/site-packages/requests/sessions.py", line 258, in post
    return self.request( post , url, data=data, **kwargs)
  File "venv/lib/python2.7/site-packages/requests/sessions.py", line 208, in request
    r.send(prefetch=prefetch)
  File "venv/lib/python2.7/site-packages/requests/models.py", line 575, in send
    timeout=self.timeout,
  File "venv/lib/python2.7/site-packages/requests/packages/urllib3/connectionpool.py", line 383, in urlopen
    body=body, headers=headers)
  File "venv/lib/python2.7/site-packages/requests/packages/urllib3/connectionpool.py", line 261, in _make_request
    httplib_response = conn.getresponse()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 1013, in getresponse
    response.begin()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 402, in begin
    version, status, reason = self._read_status()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/httplib.py", line 360, in _read_status
    line = self.fp.readline()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/socket.py", line 430, in readline
    data = recv(1)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/ssl.py", line 219, in recv
    return self.read(buflen)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/ssl.py", line 138, in read
    return self._sslobj.read(len)
KeyboardInterrupt

是否有办法避免这种或可耻的脱离关系?

Answer 1

你们只需要追捕主牌Interrupt例外。 http://docs.python.org/library/vilions.html# Tenion-hierarchy” rel=“nofollow” 。

try:
   do_something()
except KeyboardInterrupt:
   cleanup()

这也是为什么<代码>除外<>除外>是坏账的另一个例子。它不仅会追捕你想要的东西,而且可能无法追捕你们想要的东西。明确的例外。

友情链接