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

当使用 Python 2.7和 urllib2从 API 检索数据时，我得到了错误 [Errno 104] Connection reset by peer。是什么导致了这个错误，以及如何处理这个错误以使脚本不会崩溃？

心跳加速

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
324173 次浏览
小开

"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)

So you can't do anything about it, it is the issue of the server.

But you could use try .. except block to handle that exception:

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.
小开

You can try to add some time.sleep calls to your code.

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.

In order to avoid your code from crashing, try to catch this error with try .. except around the urllib2 calls.

小开

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.

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
小开

there are 2 solution you can try.

  1. request too frequently. try sleep after per request
time.sleep(1)
  1. the server detect the request client is python, so reject. add User-Agent in header to handle this.
    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

the second solution save me


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