Package: python-pycurl
Version: 7.16.2.1-3+b1
Severity: wishlist
The 'pycurl.error' object is pretty much undocumented.
To find out how to access the error code, I had to trigger an exception and
inspect the resulting exception object...
The it's easy:
try:
c.perlform()
except pycurl.error, e:
print "Error code: ", e[0]
print "Error message: ", e[1]
(you can also use e.args[0])
IMHO this should be documented somewhere in "pydoc pycurl.error"
--- System information. ---
Architecture: i386
Kernel: Linux 2.6.21-2-686
Debian Release: lenny/sid
500 unstable ftp.de.debian.org
1 experimental ftp.de.debian.org
--- Package information. ---
Depends (Version) | Installed
===================================-+-================
libc6 (>= 2.5-5) | 2.6-2
libcurl3-gnutls (>= 7.16.2-1) | 7.16.4-1
libkrb53 (>= 1.6.dfsg.1) | 1.6.dfsg.1-6
python (<< 2.6) | 2.4.4-6
python (>= 2.4) | 2.4.4-6
best regards,
Erich Schubert
--
erich@(vitavonni.de|debian.org) -- GPG Key ID: 4B3A135C (o_
Why waste time learning, when ignorance is instantaneous? --- Calvin //\
Die kürzeste Verbindung zwischen zwei Menschen ist ein Lächeln. V_/_
