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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

 

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