Python 處理 socket.error: [Errno 104] Connection reset by peer

Question

當使用帶有urllib2的 Python 2.7 從 API 檢索數據時，我收到錯誤[Errno 104] Connection reset by peer 。 是什么導致了錯誤，應該如何處理錯誤以使腳本不會崩潰？

股票代碼.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 <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

Answer 1

“由對等方重置連接”是 TCP/IP 等價物，相當於將電話重新掛斷。 這比僅僅不回復而留下一個懸而未決更有禮貌。 但這不是真正禮貌的 TCP/IP 對話者所期望的 FIN-ACK。 （來自其他SO答案）

所以你無能為力，這是服務器的問題。

但是您可以使用try .. except塊來處理該異常：

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.

Answer 2

您可以嘗試在代碼中添加一些time.sleep調用。

服務器端似乎將每個時間單位（小時、天、秒）的請求數量限制為安全問題。 您需要猜測有多少（也許使用帶有計數器的另一個腳本？）並調整您的腳本以不超過此限制。

為了避免您的代碼崩潰，請嘗試使用try .. except urllib2 調用捕獲此錯誤。

Answer 3

有一種方法可以使用ConnectionResetError 直接在except 子句中捕獲錯誤，更好地隔離正確的錯誤。 此示例還捕獲超時。

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

Answer 4

您可以嘗試兩種解決方案。

1. 要求太頻繁。 每個請求后嘗試sleep

time.sleep(1)

2. 服務器檢測到請求客戶端是python，所以拒絕。 在 header 中添加User-Agent來處理這個問題。

    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

第二種解決方案救了我