httplib slow read
setattr at yahoo.no
Fri Dec 7 10:24:26 CET 2001
John Hunter <jdhunter at nitace.bsd.uchicago.edu> wrote in message news:<m2u1v4w55n.fsf at mother.paradise.lost>...
> >>>>> "Kjetil" == Kjetil Jacobsen <setattr at yahoo.no> writes:
> Kjetil> another option may be to use the pycurl module which wraps
> Kjetil> the curl library:
> Thanks for the tip. I have installed it and it looks nice. One thing
> I have not been able to figure out is how to direct pycurl to print
> the HTTP header it would send were I too invoke 'perform'.
> Something like
> import pycurl
> c = pycurl.init()
> c.setopt(pycurl.URL, 'http://www.python.org')
> c.setopt(pycurl.REFERER, 'http://www.yahoo.com)
> # tried this but it ain't right.
> print pycurl.HTTPHEADER
> How can I do this?
pycurl.HTTPHEADER is used to set additional headers which will be used
when doing the request. see e.g. tests/basicfirst.py on how to use
the HTTPHEADER option to set custom headers.
afaik there is no way to get the header _before_ doing the actual
however, you can turn on the verbose flag:
which will print the header when perform() is invoked.
you may also use the pycurl.CUSTOMREQUEST to hand-craft your own
header to be used when doing the perform().
> Aside from the internal __doc__ and the cURL pages and the
> pycurl/tests files, is there any documentation I should be aware of?
pycurl closely follows the behaviour of libcurl. just remove the
CURLOPT_ prefix for setopt() and the CURLINFO_ prefix for getinfo().
i am not aware of any other documentation.
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