Pythonハンドリングsocket.error：[Errno 104]接続先のリセット

Question

urllib2でPython 2.7を使用してAPIからデータを取得すると、エラー[Errno 104] Connection reset by peerが返されます。何が原因でエラーが発生し、スクリプトがクラッシュしないようにエラーを処理する必要がありますか？

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 

トレースバック

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

"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)

だから、あなたはそれについて何もできない、それはサーバーの問題です。

しかし、あなたはその例外を処理するためにtry .. exceptブロックを使用することができます。

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

あなたのコードにいくつかのtime.sleep呼び出しを追加しようとすることができます。

セキュリティ上の問題として、サーバー側でtimeunit（時間、日、秒）ごとの要求量が制限されているようです。あなたはどれくらいの数を推測する必要がありますか（カウンタで別のスクリプトを使用するのでしょうか？）、この制限を超えないようにスクリプトを調整します。

コードがクラッシュしないようにするには、urllib2呼び出しの周りでtry .. exceptを使用してこのエラーをキャッチしてください。