I don't think your message is constructed correctly. I also don't see where it incldues the file contents itself. Rather than trying to construct the MIME-encoded POST form yourself, use the MimeWriter and urllib standard libraries. The MimeWriter docs give an example of encoding multiple files into one MIME document, and urllib has a command to send POST data in urlopen(). This are what I would use (I have used urllib to POST data, but not file upload data so far). Cheers, Tom P [Zanotti Michele]
Perhaps this is a Python question, not a Zope problem, but I hope someone can help me. I'm tryng to upload a file from PythonWin Ide (I have setted my Zope-test permissions to anonymous). I obtain a 400 error with this page: "...Invalid request id omitted...": but I don't understand where are my error(s). This is my code:
import httplib bd='--7d01ecf406a6' body=bd+'\r\ncontent-disposition: form-data; name="id"\r\n\r\naloha\r\n'+bd+'\r\ncontent-disposition: form-data; name="title"\r\n\nxx\r\n'+bd+'\r\ncontent-disposition: form-data; name="submit"\r\n\r\n add\r\n' body=body+bd+'\r\ncontent-disposition: form-data; name="file"; filename="prova.txt"\r\ncontent-type: text/plain\r\n\r\nprova\r\n'+bd+'--' conn = httplib.HTTPConnection("mypc:8080") conn.putrequest("POST", "/manage_addFile") conn.putheader('content-type','multipart/form-data; boundary='+bd) conn.putheader('content-length',str(lenb)) conn.endheaders() conn.send(body) r=conn.getresponse() r.status 400 print r.read() ...<P><STRONG>Invalid request</STRONG></P> The parameter, <em>id</em>, was omitted from the request.<p>... Can anyone help me? Thanks and excuse if my errors are so trivial... Michele P.S. perhaps a better solution for batch upload is Curl or pyCurl, but I find pycurl has too poor documentation for my level... Has anyone used it?