Not sure I follow... If you're using web2py cron I'd suggest calling the function directly (see the cron doc page on syntax). If you're using service mode on windows and this does not work out, you can still call the function directly (e.g. python web2py.py -S application/ controller/function in the crontab)
On Aug 4, 7:17 pm, Hasanat Kazmi <hasanatka...@gmail.com> wrote: > I have a website where my cron job (present in application/cron) calls > webservices hosted by localhost. I have the whole URL but except the > port at which the website is hosted. I don't want to hardcode port 80, > any idea how to do it? > > Hasanat Kazmi --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "web2py-users" group. To post to this group, send email to web2py@googlegroups.com To unsubscribe from this group, send email to web2py+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/web2py?hl=en -~----------~----~----~----~------~----~------~--~---
