Python處理socket.error：[Errno 104]由對端重置連接

當使用Python 2.7與urllib2從API檢索數據時，出現錯誤[Errno 104] Connection reset by peer。什麼導致錯誤，以及如何處理錯誤，以便腳本不會崩潰？Python處理socket.error：[Errno 104]由對端重置連接

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

回溯

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

來源

2013-12-13 Athena Wisdom

在arch linux上，get_rate對我來說工作正常。你確定你沒有被過濾嗎？你可以在瀏覽器中加載該網址嗎？ – korylprince

@korylprince它在瀏覽器中正常工作，腳本在錯誤開始出現之前運行良好。如果我無法避免這個錯誤，那麼應該如何處理這個錯誤，以便它不會崩潰，並且可能使用最新檢索到的值？ –

"Connection reset by peer" is the TCP/IP equivalent of slamming the phone back on the hook. It's more polite than merely not replying, leaving one hanging. But it's not the FIN-ACK expected of the truly polite TCP/IP converseur. (From other SO answer)

所以你不能做任何事情，它是服務器的問題。

但是你可以使用try .. except塊來處理這個異常：

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.

來源

2013-12-13 14:27:16 Bunyk

服務器的管理員通常使用這種方法阻止客戶的潛在刮取請求，還是更可能只是一個無意的錯誤？現在我想知道我是否有意阻擋... – Blaszard

您可以嘗試一些time.sleep呼叫添加到您的代碼。

似乎服務器端限制每個時間單元（小時，天，秒）的請求數量作爲安全問題。你需要猜測有多少（也許用一個計數器使用另一個腳本？），並調整你的腳本不超過這個限制。

爲了避免您的代碼崩潰，請嘗試在urllib2調用周圍使用try .. except來捕獲此錯誤。

來源

2017-06-19 15:02:40

Python處理socket.error：[Errno 104]由對端重置連接

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