Hi everybody!
I'm trying to start a cron job, which fails:
2011-05-24 10:40:00,517 - web2py.cron - WARNING - WEB2PY CRON Call
returned code 1:
web2py Enterprise Web Framework
Created by Massimo Di Pierro, Copyright 2007-2011
Version 1.95.1 (2011-04-25 15:04:14)
Database drivers available: SQLite3, pymysql, Oracle
Traceback (most recent call last):
File "/opt/ecc/web2py/web2py.py", line 19, in <module>
gluon.widget.start(cron=True)
File "/opt/ecc/web2py/gluon/widget.py", line 804, in start
import_models=options.import_models, startfile=options.run)
File "/opt/ecc/web2py/gluon/shell.py", line 203, in run
execfile(startfile, _env)
IOError: [Errno 2] No such file or directory: 'applications/ecc/cron/
ecc_importer.py'
This is my contab entry:
0-59/5 * * * * eswsys *applications/ecc/
cron/ecc_importer.py
With a logger.warning call and os.getcwd() just before the execfile I
get:
2011-05-24 10:40:00,468 - web2py - WARNING - EXECFILE: applications/
ecc/cron/ecc_importer.py, working dir: /opt/ecc/web2py
The curent working directory seems to be ok and ecc_importer.py is in
the right place with 755 rights.
The job run fine on my Windows development machine but now doesn't
work when I moved it to our Linux server (Red Hat 5.2 (Tikanga) /
CentOS, 64bit).
Any ideas what I'm doing wrong?
Do I have to use PYTHONSTARTUP somehow?
TIA -.- Marcel
