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
Re: Python XMLRPC question
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
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
On Oct 13, 2:22 pm, "Gabriel Genellina" wrote: > En Tue, 13 Oct 2009 16:55:09 -0300, Falcolas escribió: [snip] > > 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 Indeed - I should probably be more explicit that my thoughts were aimed at logging into their own file. Garrick -- http://mail.python.org/mailman/listinfo/python-list
Re: Python XMLRPC question
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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Python XMLRPC question
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. Garrick -- 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
