hello, I have a url that is "http://query.directrdr.com/ptrack? pid=225&v_url=http:// www.plentyoffish.com&keyword=flowers&feed=1&ip=12.2.2.2&said=$said". If I open it on a browser, I can get its contents without any problem. However, if I use following code,
import urllib2 url = 'http://query.directrdr.com/ptrack?pid=225&v_url=http:// www.plentyoffish.com&keyword=flowers&feed=1&ip=12.2.2.2&said=$said' xml = urllib2.urlopen(url).read() then I get an exception of File "/usr/lib/python2.5/urllib2.py", line 1082, in do_open raise URLError(err) urllib2.URLError: <urlopen error (-2, 'Name or service not known')> I think this is caused by the embedded "v_url=..." in the url. Anybody knows how to fix this? By the way, this code works well if I change the value of url to something like "www.yahoo.com" or "www.google.com". Thanks so much. -- http://mail.python.org/mailman/listinfo/python-list
