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当使用带有 urllib2
的 Python 2.7 从 API 检索数据时,我收到错误 [Errno 104] Connection reset by peer
。是什么导致了错误,应该如何处理错误以使脚本不会崩溃?
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
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
最佳答案
"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)
所以你无能为力,这是服务器的问题。
但您可以使用 try .. except
block 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.
关于Python处理socket.error : [Errno 104] Connection reset by peer,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20568216/
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