I might have been unclear, or this tid-bit might have been lost in the thread... but I'm trying to send directly from ImageGrab.Grab(), without saving the data as a file. Thats where I'm getting hung... If it try to send an actual stored file, I have no problem. Is this maybe impossible? My thought was that I could just save a little process time and file fragmentation if I cut out the middle man, plus there really is no reason to save the screen capture on the server side.
Maybe I really need to look into SOAP for this sort of stuff? I'm just playing with the technology, and from the searching I've done, the XML-RPC seemed to fit my needs best. I could certainly be wrong though. Thanks for both of you giving me feedback. On 12/29/06, Lee Harr <[EMAIL PROTECTED]> wrote:
>http://www.velocityreviews.com/forums/t343990-xmlrpc-send-file.html > >Using this example I get error's about 'expected binary .read(), but got >instance instead. I assume you are using this ... >d = xmlrpclib.Binary(open("C:\\somefile.exe").read()) Are you using windows? I think you would need to pass the binary flag to open ... imagedata = open(filename, 'rb').read() It's probably a good idea to use the binary flag if you are expecting binary data just in case it gets ported somewhere else later. >I've just been using xmlrpclib and simplexmlrpcserver for this, but I'm >wondering if I should perhaps use twisted instead. I've used xml-rpc to send image data before. It worked. _________________________________________________________________ Don't just search. Find. Check out the new MSN Search! http://search.msn.com/ _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
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