Imagine a script is running in these 2 sets of "conditions":
sudo crontab./my-script.pyWhat I'd like to achieve is an automatic detection of "debug mode", without me specifying an argument (e.g. --debug) for the script.
Is there a convention about how to do this? Is there a variable that can tell me who the script owner is? Whether script has a console at stdout? Run a ps | grep to determine that?
Thank you for your time.
Since sys.stdin will be a TTY in debug mode, you can use the os.isatty() function:
import sys, os
if os.isatty(sys.stdin.fileno()):
# Debug mode.
pass
else:
# Cron mode.
pass
You could add an environment variable to the crontab line and check, inside your python application, if the environment variable is set.
crontab's configuration file:
CRONTAB=true
# run five minutes after midnight, every day
5 0 * * * /path/to/your/pythonscript
Python code:
import os
if os.getenv('CRONTAB') == 'true':
# do your crontab things
else:
# do your debug things
Use a command line option that only cron will use.
Or a symlink to give the script a different name when called by cron. You can then use sys.argv[0]to distinguish between the two ways to call the script.
os.isatty() to detect whether sysout is a terminal is a hack. the OP specifically said he did not want to to do it with a command line argument. In addition, it sounds like Python itself might be doing something like this in certain situations -- see the info for the -i interpreter command line option in the 1.1.3. Miscellaneous options section of the online docs.