urllib2 file upload error
Thomas
thomas at eforms.co.nz
Wed Apr 27 23:45:50 EDT 2005
import urllib2
URL = 'http://xxx.xxx.xxx.xxx/UploaderHttp/'
FILE= 'c:/Documents and Settings/Administrator/Desktop/Vousleve 1 .pps';
d = [ ('cmd', 'add-item'),
('protocol_version', '2.1'),
('userfile', open(FILE)),
]
req = urllib2.Request(URL, d)
try:
u = urllib2.urlopen(req)
except urllib2.HTTPError, errobj:
print "HTTPError:", errobj.code
else:
buf = u.read()
print "OK"
Python 2.3.4 (#53, May 25 2004, 21:17:02) [MSC v.1200 32 bit (Intel)] on
win32
Type "help", "copyright", "credits" or "license" for more information.
>>> ## working on region in file
c:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/python-1396n5Q.py...
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "c:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/python-1396n5Q.py", line 11, in ?
u = urllib2.urlopen(req)
File "c:\Python23\lib\urllib2.py", line 129, in urlopen
return _opener.open(url, data)
File "c:\Python23\lib\urllib2.py", line 326, in open
'_open', req)
File "c:\Python23\lib\urllib2.py", line 306, in _call_chain
result = func(*args)
File "c:\Python23\lib\urllib2.py", line 901, in http_open
return self.do_open(httplib.HTTP, req)
File "c:\Python23\lib\urllib2.py", line 888, in do_open
h.send(data)
File "C:\Python23\lib\httplib.py", line 576, in send
self.sock.sendall(str)
File "<string>", line 1, in sendall
TypeError: sendall() argument 1 must be string or read-only buffer, not list
Tried the script from http://fabien.seisen.org/python/urllib2_multipart.html
also
Getting similar error..
Cheers
Thomas
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.python.org/pipermail/python-list/attachments/20050428/d211b5d5/attachment.html>
More information about the Python-list
mailing list