Hello everyone,
I just migrated an old (2017-ish) web2py app, which was still hosted on a
user machine, running on Rocket HTTP server.
The app is now on a CentOS 7 VM, under nginx/uwsgi using HTTPS.
Web2py version is 2.27.1, over Python 2.7.5, in a virtualenv.
All is OK as for the app itself, but now I as trying to setup cron to to
delete old sessions, and I'm stuck with this :
*$ python web2py.py -S <myapp> -M -R scripts/sessions2trash.py -A -o -X
3600 -f -v*
Traceback (most recent call last):
File "web2py.py", line 44, in <module>
import gluon.widget
File "/home/apinho/web2py/web2py/gluon/widget.py", line 28, in <module>
from gluon.console import console, is_appdir
File "/home/apinho/web2py/web2py/gluon/console.py", line 53, in <module>
from gluon.shell import die
File "/home/apinho/web2py/web2py/gluon/shell.py", line 305
exec(read_pyc(pycfile), _env)
SyntaxError: unqualified exec is not allowed in function 'run' it contains
a nested function with free variables
I really have no clue what to do. There is little help online. The only one
I could find says that upgrading to Python 2.7.18 would do the trick, but
that would most probably wreck Centos7 (Which I am stuck with, for it is my
company's mandatory Linux distro for production environments).
Any help would be much appreciated.
Alexandre
--
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)
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