...this looks like the right stuff! ---v
On Sep 23, 10:48 pm, bud <[email protected]> wrote: > 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. > > best cheers > -b -- 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.
