[Tutor] XML-RPC data transfers.

[Chris Hengge]

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 <missive at hotmail.com> 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.
>
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