url = "www.someurl.com"
request = urllib2.Request(url,header={"User-agent" : "Mozilla/5.0"})
contentString = urllib2.url(request).read()
contentFile = StringIO.StringIO(contentString)
for i in range(0,2):
html = contentFile.readline()
print html
The above code runs fine from commandline but if i add it to a cron job it throws the following error:
File "/usr/lib64/python2.6/urllib2.py", line 409, in _open
'_open', req)
File "/usr/lib64/python2.6/urllib2.py", line 369, in _call_chain
result = func(*args)
File "/usr/lib64/python2.6/urllib2.py", line 1186, in http_open
return self.do_open(httplib.HTTPConnection, req)
File "/usr/lib64/python2.6/urllib2.py", line 1161, in do_open
raise URLError(err)
urllib2.URLError:
I did look at some tips on the other forums and tried it but it has been of no use.
Any help will be much appreciated.
print sys.version; print sys.pathat the top of the program and compare 2. Compare environment variables (print os.environ). Check if aHTTP_PROXYenvironment variable is present. - codeapeprint os.environfrom the program and make sure that output is redirected so that you can inspect the value. - codeape