New Bee
Hi all, I am new to this group.I am willing to learn PYTHON can anyone suggest me where to start.I have installed PYTHON 2.4 windows platform.Where can I get books for beginners.How to go about?.It would be helpful if anyone is will to guide me. Thanks and Regards, Prasanna -- http://mail.python.org/mailman/listinfo/python-list
Re: New Bee
Thanx for ur info. Prasanna -- http://mail.python.org/mailman/listinfo/python-list
Python XMLRPC question
In using Python's XMLRPC, there is a statement that gets printed on stdout of the form: localhost - - [12/Oct/2009 23:36:12] "POST /RPC2 HTTP/ 1.0" 200 - Where does this message originate? Can I turn it off, or at least redirect it into a logging file? I am planning to run the server code automatically at start up and wouldn't have a terminal window open to get this message. I guess I could redirect/pipe it to a file, but it would be more useful I could send it to the same log file that I have the server writing other messages to. Thanks for any help. -- http://mail.python.org/mailman/listinfo/python-list
Re: Python XMLRPC question
On Oct 13, 1:22 pm, "Gabriel Genellina" wrote: > En Tue, 13 Oct 2009 16:55:09 -0300, Falcolas escribió: > > > > > > > On Oct 13, 12:47 pm,prasanna wrote: > >> In using Python's XMLRPC, there is a statement that gets printed on > >> stdout of the form: > >> localhost - - [12/Oct/2009 23:36:12] "POST /RPC2 HTTP/ > >> 1.0" 200 - > > >> Where does this message originate? Can I turn it off, or at least > >> redirect it into a logging file? I am planning to run the server code > >> automatically at start up and wouldn't have a terminal window open to > >> get this message. I guess I could redirect/pipe it to a file, but it > >> would be more useful I could send it to the same log file that I have > >> the server writing other messages to. > > >> Thanks for any help. > > > Looking back through the SimpleXMLRPCServer code, it appears that this > > happens if the logRequests parameter in the SimpleXMLRPCServer class > > initialization is set to True, which it is by default. The underlying > > implementation of the logging is in BaseHTTPServer.py, which uses > > sys.stderr. > > > Looks like the simplest way to change that would be to inherit from > > the SimpleXMLRPCRequestHandler class and implement your own > > log_request method. You could then pass that to the SimpleXMLRPCServer > > constructor. > > I think it's easier to pass logRequests=False when creating the server. > > -- > Gabriel Genellina Thanks. That helped get rid of the message. Out of curiosity--one more thing I haven't yet figured out, is there a xmlrpc command I can send that stops or restarts the server? --p -- http://mail.python.org/mailman/listinfo/python-list
Re: Python XMLRPC question
Thanks a bunch. Qill give it a shot.
--p
On Oct 14, 8:18 pm, "Gabriel Genellina"
wrote:
> En Wed, 14 Oct 2009 22:08:09 -0300,prasanna
> escribió:
>
> > Out of curiosity--one more thing I haven't yet figured out, is there a
> > xmlrpc command I can send that stops or restarts the server?
>
> If you're using Python 2.6, the easiest way is to register its shutdown()
> method. Note that it *must* be called from a separate thread (just inherit
> from ForkingMixIn)
>
> On earlier versions, overwrite the serve_forever loop (so it reads `while
> not self._quit: ...`) and add a shutdown() method that sets self._quit to
> True. You'll need to call shutdown twice in that case.
>
> === begin xmlrpcshutdown.py ===
> import sys
>
> def server():
> from SocketServer import ThreadingMixIn
> from SimpleXMLRPCServer import SimpleXMLRPCServer
>
> # ThreadingMixIn must be included when publishing
> # the shutdown method
> class MyXMLRPCServer(ThreadingMixIn, SimpleXMLRPCServer):
> pass
>
> print 'Running XML-RPC server on port 8000'
> server = MyXMLRPCServer(("localhost", 8000),
> logRequests=False, allow_none=True)
> # allow_none=True because of shutdown
> server.register_function(lambda x,y: x+y, 'add')
> server.register_function(server.shutdown)
> server.serve_forever()
>
> def client():
> from xmlrpclib import ServerProxy
>
> print 'Connecting to XML-RPC server on port 8000'
> server = ServerProxy("http://localhost:8000";)
> print "2+3=", server.add(2, 3)
> print "asking server to shut down"
> server.shutdown()
>
> if sys.argv[1]=="server": server()
> elif sys.argv[1]=="client": client()
> === end xmlrpcshutdown.py ===
>
> C:\TEMP>start python xmlrpcshutdown.py server
>
> C:\TEMP>python xmlrpcshutdown.py client
> Connecting to XML-RPC server on port 8000
> 2+3= 5
> asking server to shut down
>
> C:\TEMP>
>
> --
> Gabriel Genellina
--
http://mail.python.org/mailman/listinfo/python-list
error
Sir/mam, Fatal error in launcher: Unable to create process using this is error which I am not getting clear with it . so, I request you to help me it. -- https://mail.python.org/mailman/listinfo/python-list
