how to make a SimpleXMLRPCServer abort at CTRL-C under windows

[News123]

Hi,



I'm using an XMLRPC server under Windows.

What I wonder is how I could create a server, that can be killed with CTRL-C

The server aborts easily with CTRL-BREAK but not with CTRL-C (under Windows)

If I press CTRL-C it will only abort when the next RPC call occurs.
It seems it is blocking in the select() call in the handle_request()
function.

Is there any trick, which I overlook?

thanks in advance for ideas and bye


N

 

[Gabriel Genellina]

I'm using an XMLRPC server under Windows.

What I wonder is how I could create a server, that can be killed with
CTRL-C

The server aborts easily with CTRL-BREAK but not with CTRL-C (under
Windows)

If I press CTRL-C it will only abort when the next RPC call occurs.
It seems it is blocking in the select() call in the handle_request()
function.

Python 2.6 and up behaves exactly as you want.
On previous versions you may use this:

class MyXMLRPCServer(SimpleXMLRPCServer.SimpleXMLRPCServer):

... your methods ...

if not hasattr(SimpleXMLRPCServer.SimpleXMLRPCServer, 'shutdown'):

# pre 2.6
quit = False

def serve_forever(self):
while not self.quit:
self.handle_request()

def shutdown(self):
self.quit = True

def server_bind(self):
self.socket.settimeout(1.0)
SimpleXMLRPCServer.SimpleXMLRPCServer.server_bind(self)

 

[News123]

Hi Gabriel,

I'm using an XMLRPC server under Windows.

What I wonder is how I could create a server, that can be killed with
CTRL-C

The server aborts easily with CTRL-BREAK but not with CTRL-C (under
Windows)

If I press CTRL-C it will only abort when the next RPC call occurs.
It seems it is blocking in the select() call in the handle_request()
function.

Python 2.6 and up behaves exactly as you want.
On previous versions you may use this:]

I', using python 2.6.4 nd neither on Win-XP nor on Win-7 I'm capable to
abort with CTR-C ?


On Linux however I can use CTRL-C .

class MyXMLRPCServer(SimpleXMLRPCServer.SimpleXMLRPCServer):

... your methods ...

if not hasattr(SimpleXMLRPCServer.SimpleXMLRPCServer, 'shutdown'):

# pre 2.6
quit = False

def serve_forever(self):
while not self.quit:
self.handle_request()

def shutdown(self):
self.quit = True

def server_bind(self):
self.socket.settimeout(1.0)
SimpleXMLRPCServer.SimpleXMLRPCServer.server_bind(self)

I tried something similiar, but without setting the socket timeout.
I'll try it now with.


N

 

[News123]

Hi Gabriel,

Python 2.6 and up behaves exactly as you want.
On previous versions you may use this:

class MyXMLRPCServer(SimpleXMLRPCServer.SimpleXMLRPCServer):

... your methods ...

if not hasattr(SimpleXMLRPCServer.SimpleXMLRPCServer, 'shutdown'):

# pre 2.6
quit = False

def serve_forever(self):
while not self.quit:
self.handle_request()

def shutdown(self):
self.quit = True

def server_bind(self):
self.socket.settimeout(1.0)
SimpleXMLRPCServer.SimpleXMLRPCServer.server_bind(self)

Overloading server_bind() with your version solved my problem.

thanks again

N

 

[Gabriel Genellina]

Overloading server_bind() with your version solved my problem.

Strange. With Python 2.6.4 I don't need to do that; I'd say the difference
is in the OS or antivirus (some AV are known to break the TCP stack).

 

[Aahz]

Strange. With Python 2.6.4 I don't need to do that; I'd say the difference
is in the OS or antivirus (some AV are known to break the TCP stack).

Perhaps, but I've also found that ctrl-C doesn't work on Windows.

 

[Frank Millman]

Perhaps, but I've also found that ctrl-C doesn't work on Windows.

I don't know if this is exactly the same, but FWIW ...

I had the following, in a test program -

from wsgiref.simple_server import make_server

myserver = MyServer()
httpd = make_server('', 7789, myserver)
httpd.serve_forever()

try:
while True:
sleep(1)
except KeyboardInterrupt:
httpd.shutdown()

I could not get it to respond to Ctrl-C.

Then I read somewhere that it must run in a separate thread, so I changed it
like this -

- httpd.serve_forever()
+ threading.Thread(target=httpd.serve_forever).start()

Now it does respond to Ctrl-C.

HTH

Frank Millman

 

[Dennis Lee Bieber]

Perhaps, but I've also found that ctrl-C doesn't work on Windows.

Unless the running program makes an I/O call to the console, I don't
think <ctrl-c> gets past the device driver... <G>

 

[Aahz]

Unless the running program makes an I/O call to the console, I don't
think <ctrl-c> gets past the device driver... <G>

That's probably it. It's more annoying for me because I run Windows with
a VM on a Mac, which doesn't have ctrl-break.

 

[Gabriel Genellina]

That's probably it. It's more annoying for me because I run Windows with
a VM on a Mac, which doesn't have ctrl-break.

On a "real" PC with Windows XP SP3 and Python 2.6.4, with an idle server,
just hitting Ctrl-C was enough:

D:\temp>python -m SimpleXMLRPCServer
Running XML-RPC server on port 8000
Traceback (most recent call last):
File "d:\apps\python26\lib\runpy.py", line 122, in _run_module_as_main
"__main__", fname, loader, pkg_name)
File "d:\apps\python26\lib\runpy.py", line 34, in _run_code
exec code in run_globals
File "d:\apps\python26\lib\SimpleXMLRPCServer.py", line 615, in <module>
server.serve_forever()
File "d:\apps\python26\lib\SocketServer.py", line 224, in serve_forever
r, w, e = select.select([self], [], [], poll_interval)
KeyboardInterrupt

A slightly more realistic example: a busy, single threaded server. I had
to hit Ctrl-C twice: the first time, KeyboardInterrupt was sent as a Fault
response to the client.


import sys

def busy():
x = 50000
y = x**x
return "Ok"

if sys.argv[1]=='server':
import SimpleXMLRPCServer
server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost",
8000),logRequests=False)
server.register_function(busy)
server.serve_forever()
else:
import xmlrpclib
proxy = xmlrpclib.ServerProxy("http://localhost:8000/")
while True:
print "client", proxy.busy()


Maybe the OP's code containes a bare 'except:' clause that swallows
KeyboardInterrupt, or the server is so busy that is always executing a
function handler (and all KeyboardInterrupt become Fault and are sent as a
function response).

Mmm, perhaps that's a bug, a KeyboardInterrupt should bubble up to the
server code, not being treated as an error in computing the function
result.

 

[Dennis Lee Bieber]

On a "real" PC with Windows XP SP3 and Python 2.6.4, with an idle server,
just hitting Ctrl-C was enough:

D:\temp>python -m SimpleXMLRPCServer
Running XML-RPC server on port 8000
Traceback (most recent call last):
File "d:\apps\python26\lib\runpy.py", line 122, in _run_module_as_main
"__main__", fname, loader, pkg_name)
File "d:\apps\python26\lib\runpy.py", line 34, in _run_code
exec code in run_globals
File "d:\apps\python26\lib\SimpleXMLRPCServer.py", line 615, in <module>
server.serve_forever()
File "d:\apps\python26\lib\SocketServer.py", line 224, in serve_forever
r, w, e = select.select([self], [], [], poll_interval)
KeyboardInterrupt

Well... Okay, maybe not a full console I/O request but I suspect
code that didn't use select() might also be non-responsive -- while
select() only works with sockets on Windows, it may include a check for
<ctrl-c>... code that doesn't invoke such an event-based wait, though?

 

[News123]

Hi Gabriel,

Hi Gabriel,



Overloading server_bind() with your version solved my problem.

Overloading server_bind() makes the server killable with ctrl-C.

I noticed however, that I can't use this as solution.
If the bind function sets a time out, then it seems,

that there are regular small time windows, when connecting clients fail.
I don't have the error message any more, but will post it when I capture
the event next time.


bye

N