I have an application written in web2py framework.I have a task queue which polls for every 60 seconds. The Task queue(tqueue.py) contains a program which inserts into a database.
But when I do: python web2py.py -S app1 -i 127.0.0.1 -p 9001 -M -R applications/app1/private/tqueue.py The cmd prompts shows the following information: web2py Web FrameworkCreated by Massimo Di Pierro, Copyright 2007-2016Version 2.14.3-stable+timestamp.2016.03.26.23.02.02Database drivers available: psycopg2, pymysql, imaplib, sqlite3, pg8000, pyodbc But when i try to open the webpage: http://127.0.0.1:9001/app1/default/index It doesnot open up and says connection refused. I could see the output from tqueue.py script though. I have some print statement inside the tqueue.py script which prints to the command prompt. When I try to do only: python web2py.py I can reach the administrative interface as well as open the Homepage at http://127.0.0.1:9001/app1/default/index But my task queue becomes inactive. Kindly let me know if i am doing anything wrong. -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
