Anjanesh Lekshminarayanan wrote: > fp = urllib.urlopen(url) > data = fp.read() > > Retrieving XML data via an XML service API. > Very often network gets stuck in between. No errors / exceptions. > > CTRL+C > > File "get-xml.py", line 32, in <module> > fp = urllib.urlopen(url) > File "/usr/lib/python2.6/urllib.py", line 87, in urlopen > return opener.open(url) > File "/usr/lib/python2.6/urllib.py", line 206, in open > return getattr(self, name)(url) > File "/usr/lib/python2.6/urllib.py", line 348, in open_http > errcode, errmsg, headers = h.getreply() > File "/usr/lib/python2.6/httplib.py", line 1048, in getreply > response = self._conn.getresponse() > File "/usr/lib/python2.6/httplib.py", line 974, in getresponse > response.begin() > File "/usr/lib/python2.6/httplib.py", line 391, in begin > version, status, reason = self._read_status() > File "/usr/lib/python2.6/httplib.py", line 349, in _read_status > line = self.fp.readline() > File "/usr/lib/python2.6/socket.py", line 397, in readline > data = recv(1) > KeyboardInterrupt > > Is there I can do to try something else if its taking too long to > retrieve from the network ? Like kill previous attempt and retry ? > > Thanks > Anjanesh Lekshmnarayanan
import socket from urllib2 import urlopen # A one-hundredths of a second (0.01) timeout before socket throws # an exception to demonstrate catching the timeout. # Obviously, this you will set this greater than 0.01 in real life. socket.setdefaulttimeout(0.01) # example xml feed xml_source = "http://mlb.mlb.com/partnerxml/gen/news/rss/mlb.xml"; try: data = urlopen(xml_source) except urllib2.URLError, e: print 'URLError = ' + str(e.reason) -- http://mail.python.org/mailman/listinfo/python-list
