Python 套接字错误弹性/解决方法

Question

我有一个正在运行的脚本正在测试一系列 url 的可用性。

这是功能之一。

def checkUrl(url): # Only downloads headers, returns status code.
    p = urlparse(url)
    conn = httplib.HTTPConnection(p.netloc)
    conn.request('HEAD', p.path)
    resp = conn.getresponse()
    return resp.status

有时，VPS 会失去连接，当这种情况发生时整个脚本会崩溃。

File "/usr/lib/python2.6/httplib.py", line 914, in request
  self._send_request(method, url, body, headers)
File "/usr/lib/python2.6/httplib.py", line 951, in _send_request
  self.endheaders()
File "/usr/lib/python2.6/httplib.py", line 908, in endheaders
  self._send_output()
File "/usr/lib/python2.6/httplib.py", line 780, in _send_output
  self.send(msg)
File "/usr/lib/python2.6/httplib.py", line 739, in send
  self.connect()
File "/usr/lib/python2.6/httplib.py", line 720, in connect
  self.timeout)
File "/usr/lib/python2.6/socket.py", line 561, in create_connection
  raise error, msg
socket.error: [Errno 101] Network is unreachable

我一点也不熟悉在 python 中处理这样的错误。

当网络连接暂时丢失时，防止脚本崩溃的适当方法是什么？

---

编辑:

我最终得到了这个 - 反馈？

def checkUrl(url): # Only downloads headers, returns status code.
    try:
        p = urlparse(url)
        conn = httplib.HTTPConnection(p.netloc)
        conn.request('HEAD', p.path)
        resp = conn.getresponse()
        return resp.status
    except IOError, e:
        if e.errno == 101:
            print "Network Error"
            time.sleep(1)
            checkUrl(url)
        else:
            raise

虽然我不确定我是否完全理解 raise 的作用..

Answer 1

如果你只想处理这个Network is unreachable 101，而让其他异常抛出错误，你可以做如下例子。

from errno import ENETUNREACH

try:
    # tricky code goes here

except IOError as e:
    # an IOError exception occurred (socket.error is a subclass)
    if e.errno == ENETUNREACH:
        # now we had the error code 101, network unreachable
        do_some_recovery
    else:
        # other exceptions we reraise again
        raise