Python处理socket.error:[Errno 104]由peer重置连接

当使用Python 2.7 urllib2从API检索数据时,我收到错误[Errno 104] Connection reset by peer.什么导致错误,以及如何处理错误,以便脚本不会崩溃？

ticker.py

def urlopen(url):
    response = None
    request = urllib2.Request(url=url)
    try:
        response = urllib2.urlopen(request).read()
    except urllib2.HTTPError as err:
        print "HTTPError: {} ({})".format(url, err.code)
    except urllib2.URLError as err:
        print "URLError: {} ({})".format(url, err.reason)
    except httplib.BadStatusLine as err:
        print "BadStatusLine: {}".format(url)
    return response

def get_rate(from_currency="EUR", to_currency="USD"):
    url = "https://finance.yahoo.com/d/quotes.csv?f=sl1&s=%s%s=X" % (
        from_currency, to_currency)
    data = urlopen(url)
    if "%s%s" % (from_currency, to_currency) in data:
        return float(data.strip().split(",")[1])
    return None


counter = 0
while True:

    counter = counter + 1
    if counter==0 or counter%10:
        rateEurUsd = float(get_rate('EUR', 'USD'))

    # does more stuff here

追溯

Traceback (most recent call last):
  File "/var/www/testApp/python/ticker.py", line 71, in 
    rateEurUsd = float(get_rate('EUR', 'USD'))
  File "/var/www/testApp/python/ticker.py", line 29, in get_exchange_rate
    data = urlopen(url)
  File "/var/www/testApp/python/ticker.py", line 16, in urlopen
    response = urllib2.urlopen(request).read()
  File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 406, in open
    response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 519, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 438, in error
    result = self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 625, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "/usr/lib/python2.7/urllib2.py", line 406, in open
    response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 519, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 438, in error
    result = self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 625, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "/usr/lib/python2.7/urllib2.py", line 400, in open
    response = self._open(req, data)
  File "/usr/lib/python2.7/urllib2.py", line 418, in _open
    '_open', req)
  File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 1207, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.7/urllib2.py", line 1180, in do_open
    r = h.getresponse(buffering=True)
  File "/usr/lib/python2.7/httplib.py", line 1030, in getresponse
    response.begin()
  File "/usr/lib/python2.7/httplib.py", line 407, in begin
    version, status, reason = self._read_status()
  File "/usr/lib/python2.7/httplib.py", line 365, in _read_status
    line = self.fp.readline()
  File "/usr/lib/python2.7/socket.py", line 447, in readline
    data = self._sock.recv(self._rbufsize)
socket.error: [Errno 104] Connection reset by peer
error: Forever detected script exited with code: 1

Bunyk.. 115

"通过对等方重置连接"是TCP/IP等同于将手机重新关闭.它更礼貌而不仅仅是不回复,留下一个悬挂.但这并不是真正礼貌的TCP/IP逆转所期望的FIN-ACK.(来自其他SO回答)

所以你无法做任何事情,这是服务器的问题.

但您可以使用try .. exceptblock来处理该异常:

from socket import error as SocketError
import errno

try:
    response = urllib2.urlopen(request).read()
except SocketError as e:
    if e.errno != errno.ECONNRESET:
        raise # Not error we are looking for
    pass # Handle error here.

服务器管理员通常使用这种方法来阻止客户端潜在的抓取请求，还是更可能只是一个无意的错误？现在我想知道我是否被故意阻止... (2认同)

小智.. 10

您可以尝试向time.sleep代码添加一些调用.

似乎服务器端将每个时间单位(小时,天,秒)的请求数量限制为安全问题.您需要猜测多少(可能使用另一个带有计数器的脚本？)并调整您的脚本以不超过此限制.

为了避免代码崩溃,请尝试try .. except在urllib2调用周围捕获此错误.