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.
The environment variables that were used by crontab and from the command line were different.
I fixed this by adding */15 * * * * . $HOME/.profile; /path/to/command. 
This made the crontab to pick up enivronment variables that were specified for the system.
Urllib2 runs fine if i run the program independently but throws error when i add it to a cronjob