Running a python script encapsulated in a bash function via cron

Question

I have a python script that I've encapsulated in a bash function. I would like to be able to call this function as a cron job, but I cannot seem to get cron to execute it.
The function is as follows:

 #!/bin/bash getmail(){ local interp=/path/python3 local cmd=/path/python-script local logfile=/path/logfile if [ "$1" == "-logs" ]; then $interp $cmd >> $logfile else $interp $cmd fi } 

I've then created a script to source the function and execute it which I would like to be able to call from cron.

 #!/bin/bash source /path/getmail getmail 

I've assigned this cron-script appropiate permissions, making it executable, but cron won't run the script. I can run the python script itself via cron, but not encapsulated withn the bash function. I would just like to know why. Might it have something to do with the interpreter that cron is using? I've set the SHELL=/bin/bash in cron tab. Can anyone shed some light on this for me?

Answer 1

How did you determine that it doesn't work?
Actually, what you did should almost work.
Just two things to comment on:
1. the '#!/bin/bash/ in the file containing the function is not required and, as the file is sourced, has no functionality.
2. The "-logs" parameter will never make it into the "getmail()" function since you call the function (from your script) without any parameters.
Try again with the following modification to your bash script:

#!/bin/bash source /path/getmail getmail $@ 

This should 'forward' all parameters given to your bash script into the function.