Python 处理 socket.error：[Errno 104] 对等方重置连接

问题描述：

当使用 Python 2.7urllib2从 API 检索数据时，我收到错误[Errno 104] Connection reset by peer。是什么导致了错误，以及应如何处理错误以使脚本不会崩溃？

股票行情

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 <module>
    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

解决方案 1：

“对端重置连接”相当于 TCP/IP 中将电话挂断。这比仅仅不回复而让对方挂断电话要礼貌得多。但这并不是真正有礼貌的 TCP/IP 对话者所期望的 FIN-ACK。（来自其他 SO 答案）

所以你对此无能为力，这是服务器的问题。

但您可以使用try .. except块来处理该异常：

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.

解决方案 2：

time.sleep您可以尝试在您的代码中添加一些调用。

看起来服务器端出于安全考虑限制了每个时间单位（小时、天、秒）的请求数量。您需要猜测有多少（也许使用另一个带有计数器的脚本？）并调整脚本以不超过此限制。

为了避免代码崩溃，请尝试在try .. excepturllib2 调用周围捕获此错误。

解决方案 3：

有一种方法可以直接在 except 子句中使用 ConnectionResetError 来捕获错误，更好地隔离正确的错误。此示例还捕获了超时。

from urllib.request import urlopen 
from socket import timeout

url = "http://......"
try: 
    string = urlopen(url, timeout=5).read()
except ConnectionResetError:
    print("==> ConnectionResetError")
    pass
except timeout: 
    print("==> Timeout")
    pass

解决方案 4：

有两种解决方案您可以尝试。

1. 请求过于频繁。sleep每个请求后再尝试

time.sleep(1)

1. 服务器检测到请求客户端是 python，因此拒绝。添加User-Agent标题来处理这个问题。

    headers = {
        "Content-Type": "application/json;charset=UTF-8",
        "User-Agent": "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko)"
    }
    try:
        res = requests.post("url", json=req, headers=headers)
    except Exception as e:
        print(e)
        pass

第二个解决方案拯救了我