view server.py @ 97:d1f7c1d38abe v2

success callback is no longer called if bugzilla error occurs on Bugzilla.ajax()
author Atul Varma <avarma@mozilla.com>
date Mon, 26 Apr 2010 22:16:27 -0700
parents 392fd25a6e21
children
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import os
import BaseHTTPServer
import SimpleHTTPServer

PORT = 8000

def run(server_class=BaseHTTPServer.HTTPServer,
        handler_class=SimpleHTTPServer.SimpleHTTPRequestHandler,
        port=PORT):
    server_address = ('', port)
    print "Serving files in '%s' on port %d." % (os.getcwd(), port)
    httpd = server_class(server_address, handler_class)
    httpd.serve_forever()

if __name__ == '__main__':
    run()