CODEWITHSUNDEEP
pythonxml-rpcxmlrpclib
Alex
Alex

Reputation: 44345

How can I create my custom xmlrpc fault error in python

When using xmlrpclib in python an error on the server side is reported by the client side as an xmlrpclib.Fault. A division by zero error in a method on the server side (using SimpleXMLRPCServer) usually gives an output like the following: 

xmlrpclib.Fault: <Fault 1: "<type 'exceptions.ZeroDivisionError'>:integer division or modulo by zero">

This is useful as it notes the type of error, but now where it happened. How is it possible to overwrite/modify the xmlrpclib.Fault method (in SimpleXMLRPCServer?) so it reports the error e.g. as follows: 

xmlrpclib.Fault: <Fault 1: "<type 'exceptions.ZeroDivisionError'>:integer division or modulo by zero MODULE: myMethod.py  LINE: 657">

i.e. to include the name of the module the error appeared and the line number. Is that possible to do on the server-side, without raising custom exceptions? ANY error should reported in the same way on the client side.

Upvotes: 3

Views: 1726

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1122222

If you use SimpleXMLRPCServer, you can override the internal _marshaled_dispatch method to add information to the Fault() instance generated:

This is the original method:

def _marshaled_dispatch(self, data, dispatch_method = None, path = None):
    try:
        params, method = xmlrpclib.loads(data)

        # generate response
        if dispatch_method is not None:
            response = dispatch_method(method, params)
        else:
            response = self._dispatch(method, params)
        # wrap response in a singleton tuple
        response = (response,)
        response = xmlrpclib.dumps(response, methodresponse=1,
                                   allow_none=self.allow_none, encoding=self.encoding)
    except:
        # report low level exception back to server
        # (each dispatcher should have handled their own
        # exceptions)
        exc_type, exc_value = sys.exc_info()[:2]
        response = xmlrpclib.dumps(
            xmlrpclib.Fault(1, "%s:%s" % (exc_type, exc_value)),
            encoding=self.encoding, allow_none=self.allow_none)
    return response

You can subclass SimpleXMLRPCServer.SimpleXMLRPCServer and override this method:

import SimpleXMLRPCServer
import sys
import xmlrbclib

class VerboseFaultXMLRPCServer(SimpleXMLRPCServer.SimpleXMLRPCServer):
    def _marshaled_dispatch(self, data, dispatch_method = None, path = None):
        try:
            params, method = xmlrpclib.loads(data)

            # generate response
            if dispatch_method is not None:
                response = dispatch_method(method, params)
            else:
                response = self._dispatch(method, params)
            # wrap response in a singleton tuple
            response = (response,)
            response = xmlrpclib.dumps(response, methodresponse=1,
                                       allow_none=self.allow_none, encoding=self.encoding)
        except:
            # report low level exception back to server
            # (each dispatcher should have handled their own
            # exceptions)
            exc_type, exc_value, tb = sys.exc_info()
            while tb.tb_next is not None:
                tb = tb.tb_next  # find last frame of the traceback
            lineno = tb.tb_lineno
            code = tb.tb_frame.f_code
            filename = code.co_filename
            name = code.co_name
            response = xmlrpclib.dumps(
                xmlrpclib.Fault(1, "%s:%s FILENAME: %s LINE: %s NAME: %s" % (
                    exc_type, exc_value, filename, lineno, name)),
                encoding=self.encoding, allow_none=self.allow_none)
        return response

Then use VerboseFaultXMLRPCServer instead of SimpleXMLRPCServer.SimpleXMLRPCServer.

Upvotes: 4

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