Question

I have created a simple Python XML-RPC implementation, largely based on the examples.

However, it sends output like this:

foo.bar.com - - [13/Feb/2010 17:55:47] "POST /RPC2 HTTP/1.0" 200 -

… to the terminal, even if I redirect standard out and standard error to a file using >> or >. I’m doing this with the following line:

python foobar 2>&1 >> foobar.log

It seems almost like it’s not sending to standard out, but somewhere else.

Also, when an exception occurs on recieving a request, the whole application crashes with this error:

----------------------------------------
Exception happened during processing of request from ('1.2.3.4', 51284)

How can I handle this exception? I need to recover gracefully, and just log the exception message rather than the server crashing.

Answer 1

I guess you’re using the SimpleXMLRPCServer class from the examples. In that case, simply provide the parameter logRequests when creating it:

server = SimpleXMLRPCServer(("localhost", 8000), logRequests = False)

That will suppress request logging.

As for the exceptions, they’re logged in BaseServer (cf. source code of “SocketServer.py”):

def handle_error(self, request, client_address):
    """Handle an error gracefully.  May be overridden.

    The default is to print a traceback and continue.

    """
    print '-'*40
    print 'Exception happened during processing of request from',
    print client_address
    import traceback
    traceback.print_exc() # XXX But this goes to stderr!
    print '-'*40

As you can see, the first part is written to stdout, which is why &2>1 didn’t work completely. If you want to suppress them, override or overwrite that method.