How can I make a "keep alive" HTTP request using Python's urllib2?
+1
A:
This handler might work. Look to the main introductory comment for usage details.
Oli
2009-06-24 09:55:58
For the Monty Python fans... 'E's not pinin'! 'E's passed on! This link is no more! He has ceased to be! 'E's expired and gone to meet 'is maker! 'E's a stiff! Bereft of life, 'e rests in peace! If you hadn't nailed 'im to the perch 'e'd be pushing up the daisies! 'Is metabolic processes are now 'istory! 'E's off the twig! 'E's kicked the bucket, 'e's shuffled off 'is mortal coil, run down the curtain and joined the bleedin' choir invisibile!! THIS IS AN EX-LINK!!
Marc Gravell
2010-01-05 21:30:11
Fixed the link.
Oli
2010-01-07 12:43:00
+10
A:
Use the urlgrabber library. This includes an HTTP handler for urllib2 that supports HTTP 1.1 and keepalive:
>>> import urllib2
>>> from keepalive import HTTPHandler
>>> keepalive_handler = HTTPHandler()
>>> opener = urllib2.build_opener(keepalive_handler)
>>> urllib2.install_opener(opener)
>>>
>>> fo = urllib2.urlopen('http://www.python.org')
msanders
2009-06-24 09:56:44
+1
A:
Note that urlgrabber does not entirely work with python 2.6. I fixed the issues (I think) by making the following modifications in keepalive.py.
In keepalive.HTTPHandler.do_open() remove this
if r.status == 200 or not HANDLE_ERRORS:
return r
And insert this
if r.status == 200 or not HANDLE_ERRORS:
# [speedplane] Must return an adinfourl object
resp = urllib2.addinfourl(r, r.msg, req.get_full_url())
resp.code = r.status
resp.msg = r.reason
return resp;
speedplane
2010-01-08 00:05:38