即使设置了超时，Python urllib2.urlopen也会无限冻结脚本

Question

函数 urllib2.urlopen 冻结。所以我的问题很简单：

• 为什么即使设置了超时，urlopen 也会永远冻结我的脚本？
• 如何访问 URL 上的数据（在本例中：http://api.own3d.tv /live?channel=FnaticTV) 我的 Python 进程不会永远冻结吗？

这是冻结的部分（在 own3d.py 中）：

# Try three times to make contact
while True:
    try:
        # Connect to API 

        # Right here! It freezes here
        connection = urllib2.urlopen(request, timeout=10)

        xmlstring = connection.read() 
    except URLError as e:
        tries += 1
        if tries >= 3:
            sys.stderr.write(
                      'own3dStreamsUpdater: Fatal error: Repeated timeouts')
            exit()

后的堆栈跟踪

Traceback (most recent call last):
  File "", line 1, in 
  File "honsapp/own3dStreamsUpdater.py", line 53, in updateStreamInfo
    streamInfo = getStreamInfo(stream)
  File "honsapp/own3d.py", line 98, in getStreamInfo
    connection = urllib2.urlopen(request, timeout=10)
  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 394, in open
    response = self._open(req, data)
  File "/usr/local/lib/python2.7/urllib2.py", line 412, in _open
    '_open', req)
  File "/usr/local/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/usr/local/lib/python2.7/urllib2.py", line 1199, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/local/lib/python2.7/urllib2.py", line 1170, in do_open
    r = h.getresponse(buffering=True)
  File "/usr/local/lib/python2.7/httplib.py", line 1027, in getresponse
    response.begin()
  File "/usr/local/lib/python2.7/httplib.py", line 407, in begin
    version, status, reason = self._read_status()
  File "/usr/local/lib/python2.7/httplib.py", line 365, in _read_status
    line = self.fp.readline()
  File "/usr/local/lib/python2.7/socket.py", line 447, in readline
    data = self._sock.recv(self._rbufsize)
KeyboardInterrupt

---

编辑

这是我的KeyboardInterrupt 让我的脚本运行一夜。我不知道到底花了多长时间（虽然超过五分钟）但脚本最终放弃并给了我一个堆栈跟踪：

Traceback (most recent call last):
  File "honsapp/own3dStreamsUpdater.py", line 260, in 
    newInfo()
  File "honsapp/own3dStreamsUpdater.py", line 172, in newInfo
    result = updateStreamInfo(stream)
  File "honsapp/own3dStreamsUpdater.py", line 53, in updateStreamInfo
    streamInfo = getStreamInfo(stream)
  File "/root/Dropbox/Projects/honstreams/honsapp/own3d.py", line 98, in getStreamInfo
    connection = urllib2.urlopen(request, timeout=10)
  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 394, in open
    response = self._open(req, data)
  File "/usr/local/lib/python2.7/urllib2.py", line 412, in _open
    '_open', req)
  File "/usr/local/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/usr/local/lib/python2.7/urllib2.py", line 1199, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/local/lib/python2.7/urllib2.py", line 1170, in do_open
    r = h.getresponse(buffering=True)
  File "/usr/local/lib/python2.7/httplib.py", line 1027, in getresponse
    response.begin()
  File "/usr/local/lib/python2.7/httplib.py", line 407, in begin
    version, status, reason = self._read_status()
  File "/usr/local/lib/python2.7/httplib.py", line 371, in _read_status
    raise BadStatusLine(line)
httplib.BadStatusLine: ''

Answer 1

这个脚本根本不会冻结，但它是一个很好的例子，说明了为什么你不应该在疲倦时编码。应该尝试连接到 API 3 次的循环将永远持续下去，因为我忘记在其中放置 break。

这个问题太愚蠢了，我不会责怪你删除它。

固定代码：

# Try three times to make contact
while True:
    try:
        # Connect to API 
        connection = urllib2.urlopen(request, timeout=10)
        xmlstring = connection.read()
        break
    except URLError as e:
        tries += 1
        if tries >= 3:
            sys.stderr.write(
                      'own3dStreamsUpdater: Fatal error: Repeated timeouts')
            exit()

Answer 2

您确定 urlopen() 调用挂起吗？因为如果调用成功，您的 while 循环不会终止。