If it won't be simple, it simply won't be. [Hire me, source code] by Miki Tebeka, CEO, 353Solutions

Wednesday, April 30, 2008

XML RPC File Server

#!/usr/bin/env python
'''Simple file client/server using XML RPC'''

from SimpleXMLRPCServer import SimpleXMLRPCServer
from xmlrpclib import ServerProxy, Error as XMLRPCError
import socket

def get_file(filename):
fo = open(filename, "rb")
try: # When will "with" be here?
return fo.read()
finally:
fo.close()

def main(argv=None):
if argv is None:
import sys
argv = sys.argv

default_port = "3030"
from optparse import OptionParser

parser = OptionParser("usage: %prog [options] [[HOST:]PORT]")
parser.add_option("--get", help="get file", dest="filename",
action="store", default="")

opts, args = parser.parse_args(argv[1:])
if len(args) not in (0, 1):
parser.error("wrong number of arguments") # Will exit

if args:
port = args[0]
else:
port = default_port

if ":" in port:
host, port = port.split(":")
else:
host = "localhost"

try:
port = int(port)
except ValueError:
raise SystemExit("error: bad port - %s" % port)

if opts.filename:
try:
proxy = ServerProxy("http://%s:%s" % (host, port))
print proxy.get_file(opts.filename)
raise SystemExit
except XMLRPCError, e:
error = "error: can't get %s (%s)" % (opts.filename, e.faultString)
raise SystemExit(error)
except socket.error, e:
raise SystemExit("error: can't connect (%s)" % e)

server = SimpleXMLRPCServer(("localhost", port))
server.register_function(get_file)
print "Serving files on port %d" % port
server.serve_forever()

if __name__ == "__main__":
main()


This is a huge security hole, use at your own risk.
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