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import urllib2 URL = '' 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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