Creating your own WSGI server and call start() and stop() on it, instead of using app.run(). Have a look at runsimple() in httpserver.py (part of web.py). Skeleton code:
app = web.application(urls, globals()) wsgifunc = app.wsgifunc() wsgifunc = web.httpserver.StaticMiddleware(wsgifunc) ## Enable / static webserver = web.httpserver.WSGIServer(address, wsgifunc) webserver.start() Then to shut it down do a webserver.stop() Dave On Jun 23, 3:49 pm, Graham Dumpleton <[email protected]> wrote: > On Jun 23, 11:10 am, Anand Chitipothu <[email protected]> wrote: > > > 2010/6/23 elpanson <[email protected]>: > > > > I am using embedded python and using web.py as a lightweight web > > > server. I need a way to signal it to shutdown, my application is a > > > Windows Service and does not have Keyboard interrupts. > > > Did you try calling sys.exit()? > > Usually that would not work in an embedded system as all it does is > throw a SystemExit exception. It would need the C code which calls > into the Python code to detect that type of exception and call C > exit() function or otherwise shutdown the process in an orderly > manner. > > Graham -- You received this message because you are subscribed to the Google Groups "web.py" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/webpy?hl=en.
