Hi Massimo,
At least in Windows platform, when one "web2py -i 0.0.0.0 -p 8000"
process in running, I noticed that local administrator can still start
another "web2py -i 0.0.0.0 -p 8000" process without any error message,
but of course the second web2py process can not accept any request. So
I suggest to add some detect code at the beginning of web2py
launching. Something like this:
def __main__():
# parse commandline and get the bindIP and bindPort, then
detectAddr = '127.0.0.1' if bindIP=='0.0.0.0' else bindIP
try: urllib.urlopen('http://%s:%s'% (detectAddr, bindPort)).read()
except IOError: pass # the bindAddr and bindPort is vacant
else: sys.exit('%s:%s is occupied'%(detectAddr,bindPort))
...... # Go on to start web2py main process
This code can let novice make fewer mistakes.
Regards,
Iceberg
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups
"web2py Web Framework" 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
-~----------~----~----~----~------~----~------~--~---
