bash - Cron job does NOT get the environment variables set in .bashrc

Question

Here is my cron job:

plee@dragon:~$ crontab -l
* * * * * /bin/bash -l -c 'source ~/.bashrc; echo $EDITOR > /tmp/cronjob.test'

and inside ~/.bashrc file, I have export EDITOR=vim, but in the final /tmp/cronjob.test file, it's still empty?

So how can I get the environment variables (set in .bashrc file) and use it in my cron job?

plee@dragon:~$ lsb_release -a
No LSB modules are available.
Distributor ID: Ubuntu
Description:    Ubuntu 12.04 LTS
Release:        12.04
Codename:       precise
plee@dragon:~$ uname -a
Linux dragon 3.2.0-26-generic-pae #41-Ubuntu SMP Thu Jun 14 16:45:14 UTC 2012 i686 i686 i386 GNU/Linux

If use this:

* * * * * /bin/bash -l -c -x 'source ~/.bashrc; echo $EDITOR > /tmp/cronjob.test' 2> /tmp/cron.debug.res

In /tmp/cron.debug.res:

...
++ return 0
+ source /home/plee/.bashrc
++ '[' -z '' ']'
++ return
+ echo

BTW, the .bashrc file is the default one came with Ubuntu 12.04, with the exception that I added one line export EDITOR=vim.

If I don't use the cron job, instead, just directly do this on the command line:

source .bashrc; echo $EDITOR # Output: vim

Answer 1

The reason for source ~/.bashrc not working is the contents on your ~/.bashrc (default one from Ubuntu 12.04). If you look in it you will see on lines 5 and 6 the following:

# If not running interactively, don't do anything
[ -z "$PS1" ] && return

PS1 variable is set for an interactive shell, so it's absent when run via cron, even though you are executing it as a login shell. This is confirmed by contents of the file produced by /bin/bash -l -c -x 'source ~/.bashrc; echo $EDITOR > /tmp/cronjob.test':

+ source /home/plee/.bashrc
++ '[' -z '' ']'
++ return

To make source ~/.bashrc work, comment out the line that checks for presence of the PS1 variable in ~/.bashrc:

#[ -z "$PS1" ] && return

This will make bash execute the entire contents of ~/.bashrc via cron