Python 处理 socket.error: [Errno 104] 对等方重置连接
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Python handling socket.error: [Errno 104] Connection reset by peer
提问by Athena Wisdom
When using Python 2.7 with urllib2to retrieve data from an API, I get the error [Errno 104] Connection reset by peer. Whats causing the error, and how should the error be handled so that the script does not crash?
使用 Python 2.7 withurllib2从 API 检索数据时，出现错误[Errno 104] Connection reset by peer. 导致错误的原因是什么，应该如何处理错误以便脚本不会崩溃？
ticker.py
股票代码
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
Traceback
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
回答by Bunyk
"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)
“由对等方重置连接”是 TCP/IP 等价物，相当于将电话重新挂断。这比仅仅不回复而留下一个悬而未决更有礼貌。但这不是真正礼貌的 TCP/IP 对话者所期望的 FIN-ACK。（来自其他SO答案）
So you can't do anything about it, it is the issue of the server.
所以你无能为力，这是服务器的问题。
But you could use try .. exceptblock to handle that exception:
但是您可以使用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.
回答by Felix Martinez
You can try to add some time.sleepcalls to your code.
您可以尝试time.sleep在代码中添加一些调用。
It seems like the server side limits the amount of requests per timeunit (hour, day, second) as a security issue. You need to guess how many (maybe using another script with a counter?) and adjust your script to not surpass this limit.
服务器端似乎将每个时间单位（小时、天、秒）的请求数量限制为安全问题。您需要猜测有多少（也许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。
In order to avoid your code from crashing, try to catch this error with try .. exceptaround the urllib2 calls.
为了避免您的代码崩溃，请尝试使用try .. excepturllib2 调用来捕获此错误。
回答by Cyril
There is a way to catch the error directly in the except clause with ConnectionResetError, better to isolate the right error. This example also catches the timeout.
有一种方法可以使用ConnectionResetError 直接在except 子句中捕获错误，更好地隔离正确的错误。此示例还捕获超时。
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
