2010/9/24 bud <[email protected]>:
> On Sep 23, 6:19 pm, Greg Milby <[email protected]> wrote:
>> are you trying to call the same file twice?
>
> Nope, don't even know how I could do this...
>
> The environment in which I use web.py is a daemon with multiple
> processes manages via the "multiprocessing" module. Many different
> kinds of workers work on different queues. One of the workers is a
> HTTP-REST interface to submit jobs to the initial queue. That's what
> I use web.py for... So from the command line, I launch the daemon
> process that spans and supervises all workers, including web.py...
>
> I looked through the source and found a solution:
>
> import web
> class MyApplication(web.application):
> def run(self, port=8080, *middleware):
> func = self.wsgifunc(*middleware)
> return web.httpserver.runsimple(func, ('0.0.0.0', port))
> #--------------------------
> urls = ('/', 'hello')
> class hello:
> def GET(self):
> return 'Hello'
>
> if __name__ == "__main__":
> app = MyApplication(urls, globals())
> app.run(port=8888)
>
> Since I don't need cgi etc., that works for me. Redefining "run" in a
> subclass seemed to be the easiest way..
>
> In case you like the optional "port" argument in application.run, I
> could possibly prepare a patch.. Let me know.
How about this?
if __name__ == "__main__":
web.httpserver.runsimple(app.wsgifunc(), ("0.0.0.0", 8888))
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