Create a shell script for a cron job with hidden debug information shown only when executed inside the terminal.
The idea behind this is quite simple as we need to verify if stdin (file descriptor 0), stdout (file descriptor 1), or stderr (file descriptor 2) is associated with the terminal. If not, then the shell script is executed as a cron job, so any debug message can be omitted.
#!/bin/bash
# cron job template with debug information
# display debug information
# only if $terminal variable is set to 1
emit() {
if [ "$terminal" -eq "1" ]; then
printf "%s\n" "$*"
fi
}
# determine if standard input (file descriptor) is opened on a terminal
# and set $terminal variable accordingly
if [ -t 0 ] ; then
terminal=1
else
terminal=0
fi
emit "Script name is $(basename $0)"
emit "Script location is $(readlink -f $0)"
emit "Script is executed on $TERM"
emit "Script is executed from $PWD directory"
emit "Script is executed with $USER permissions"
if [ -n "$SUDO_USER" ]; then
emit "Script is executed using sudo by $SUDO_USER ($SUDO_UID:$SUDO_GID)"
fi
emit
emit "Script is using path defined as $PATH"
emit "Script LANG variable is $(locale | awk -F '=' '/LANG=/ {print $2}')"
emit
emit "Execution start: $(date)"
# perform real operations here
emit "Execution end: $(date)"
Sample output inside the terminal.
~# bash bin/cron_job_template.sh
Script name is cron_job_template.sh Script location is /root/bin/cron_job_template.sh Script is executed on screen Script is executed from /root directory Script is executed with root permissions Script is using path defined as /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin Script LANG variable is C.UTF-8 Execution start: Tue Dec 26 13:25:50 UTC 2017 Execution end: Tue Dec 26 13:25:50 UTC 2017