Hello, I am new to web2py. I downloaded the version 1.72.3. I have an error directly at startup.
I do 'python web2py.py' to start the application server at the command line. I see this error page at http://127.0.0.1:8000/welcome/default/index: Internal error Ticket issued: welcome/127.0.0.1.2009-12-23.21-04-59.627a2785- d15a-4911-818f-8b738f9b3732. I have Python 2.6.4 running on OpenSuse 11.2 x86_64. Is it the correct way to start the server? Do I need to build something before? ------------- Pierre -- You received this message because you are subscribed to the Google Groups "web2py-users" 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/web2py?hl=en.
