Working with files in a SimpleXMLRPCServver

[Jose Carlos Balderas Alberico]

I'm setting up a server accepting XML-RPC calls using the SimpleXMLRPCServer
class. Basically, what I have to do is send a zip-compressed file to the
server, have the server unzip it and process it, after processing it the
server is supposed to zip the file again, and send it back to the client.

I found a solution, but wanted to ask you if you think there's an easier way
of doing this.

What I do at the client first is to compress the file (which is a .txt)
using os.system("zip blah blah.txt"). Then I obtain the name of the
compressed file (let's assume it is blah.zip). Once I have the name of the
zip file, I do the following:

*fd = open("blah.zip", 'r')
bin = xmlrpclib.Binary(fd.read())*
*server.process(bin)*

I obtain a file descriptor, read the data and stuff it in a Binary object,
and send it to the server as a parameter to the process method.

Then, at the server script, I create a file, and write into it the data
contained in that Binary object using the attribute .data. That way I'll
have a new file with the data passed to the server. Then again I use
os.system(...) to unzip the data and recover the original content.

When I need to send content back to the client, the whole process is
repeated.

My question is: do you think this is an appropiate way to exchange files
between client and server? I'm relativately new to Python, and the task of
the file exchange has been assigned to me. I haven't been able to find
documentation on the subject, so any help would be appreciated.

If you need any more information about what I'm trying to do, just ask.

Thank you so much for your attention :)
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