getting debug from urllib2

[Ben Edwards]

Have been experimenting with HTTP stuff in python 2.4 and am having a
problem getting debug info. If I use utllib.utlopen I get debug but if I
user utllib2 I do not. Below is the probram and the output I am
getting.

Any insight?
Ben

* Code *

import urllib, urllib2, httplib

url = 'http://www.mozillazine .org/atom.xml'
httplib.HTTPCon nection.debugle vel = 1

print "urllib"

data = urllib.urlopen( url);

print "urllib2"

request = urllib2.Request (url)
opener = urllib2.build_o pener()
feeddata = opener.open(req uest).read()

print "End\n"

* Output *

urllib
connect: (www.mozillazine.org, 80)
send: 'GET /atom.xml HTTP/1.0\r\nHost: www.mozillazine.org\r
\nUser-agent: Python-urllib/1.16\r\n\r\n'
reply: 'HTTP/1.0 200 OK\r\n'
header: Date: Mon, 31 Jul 2006 13:43:11 GMT
header: Server: Apache/2.0.55 (Gentoo) PHP/4.4.0-pl1-gentoo
header: Last-Modified: Thu, 27 Jul 2006 18:05:52 GMT
header: ETag: "20a1b4-6bcf-be12000"
header: Accept-Ranges: bytes
header: Content-Length: 27599
header: Content-Type: application/xml
header: Age: 5
header: X-Cache: HIT from mz5.mz.osuosl.o rg
header: X-Cache-Lookup: HIT from mz5.mz.osuosl.o rg:80
header: Connection: close
urllib2
End

[Peter Otten]

Re: getting debug from urllib2

Ben Edwards wrote:

Have been experimenting with HTTP stuff in python 2.4 and am having a
problem getting debug info. If I use utllib.utlopen I get debug but if I
user utllib2 I do not. Below is the probram and the output I am
getting.
>
Any insight?

Use the source :-)

urllib2.build_o pener() accepts prebuilt handlers:

import urllib2

url = 'http://www.mozillazine .org/atom.xml'

request = urllib2.Request (url)
opener = urllib2.build_o pener(urllib2.H TTPHandler(debu glevel=1))
feeddata = opener.open(req uest).read()

Peter