How to redirect urllib2 debug output without globally redefining sys.stderr?

Question

In urllib2 how can I redirect the output a particular opener generates when I specify the debuglevel parameter?

E.g. for this example

opener = urllib2.build_opener(urllib2.HTTPHandler(debuglevel=2))
opener.open("http://stackoverflow.com")

how would I catch the output in an "object" similar to sys.stderr and be able to write it anywhere I like?

I have a script that uses logging and stderr and this particular output should go to the log without hampering any of the other output.

note: [the code shows that the output is printed to `sys.stdout`, not `sys.stderr`](https://hg.python.org/cpython/file/2.6/Lib/httplib.py#l405). — jfs, Oct 30 '14 at 17:23
are there multiple threads in your program that may print while urllib2 makes requests? — jfs, Oct 30 '14 at 17:48
@J.F.Sebastian: yep, unfortunately. Thanks for your answer and your time. I guess I'll have to use locking or a separate log file for the `urllib2` debugging. — 0xC0000022L, Oct 30 '14 at 21:41

score 1 · Answer 1 · edited May 23 '17 at 10:33

1

It is easy to redirect sys.stdout temporarily for the duration of the urllib2 request but the effects are global i.e., if there are other Python threads are runnings then they'll see the change if they print something to sys.stdout.

You probably can't override what print statement does in Python 2.6. You can do it in Python 3 where print() is a function.

edited May 23 '17 at 10:33

Community

answered Oct 30 '14 at 17:28

jfs

Would `from __future__ import print_function` fix the limitation you mention? – 0xC0000022L Oct 30 '14 at 17:39
1

no. `__future__` affects only the current module. It has no effect on `urllib2` – jfs Oct 30 '14 at 17:46

1 Answers1