Databases, Systems & Networks

I ended up writing a script (I’m trying to teach myself the finer points of bash scripting, so that’s why you don’t see something like Perl here). It’s not exactly a simple affair, but it does most of what I need. It uses Kyle’s suggestion for looking up individual users’ crontabs, but also deals with /etc/crontab(including the scripts launched by run-parts in /etc/cron.hourly, /etc/cron.daily, etc.) and the jobs in the /etc/cron.d directory.source: How do I list all cron jobs for all users?

I ended up writing a script (I’m trying to teach myself the finer points of bash scripting, so that’s why you don’t see something like Perl here). It’s not exactly a simple affair, but it does most of what I need. It uses Kyle’s suggestion for looking up individual users’ crontabs, but also deals with /etc/crontab(including the scripts launched by run-parts in /etc/cron.hourly, /etc/cron.daily, etc.) and the jobs in the /etc/cron.d directory.

It takes all of those and merges them into a display something like the following:

#!/bin/bash
# System-wide crontab file and cron job directory. Change these for your system.
 CRONTAB='/etc/crontab'
 CRONDIR='/etc/cron.d'
# Single tab character. Annoyingly necessary.
 tab=$(echo -en "\t")
# Given a stream of crontab lines, exclude non-cron job lines, replace
 # whitespace characters with a single space, and remove any spaces from the
 # beginning of each line.
 function clean_cron_lines() {
 while read line ; do
 echo "${line}" |
 egrep --invert-match '^($|\s*#|\s*[[:alnum:]_]+=)' |
 sed --regexp-extended "s/\s+/ /g" |
 sed --regexp-extended "s/^ //"
 done;
 }
# Given a stream of cleaned crontab lines, echo any that don't include the
 # run-parts command, and for those that do, show each job file in the run-parts
 # directory as if it were scheduled explicitly.
 function lookup_run_parts() {
 while read line ; do
 match=$(echo "${line}" | egrep -o 'run-parts (-{1,2}\S+ )*\S+')
if [[ -z "${match}" ]] ; then
 echo "${line}"
 else
 cron_fields=$(echo "${line}" | cut -f1-6 -d' ')
 cron_job_dir=$(echo "${match}" | awk '{print $NF}')
if [[ -d "${cron_job_dir}" ]] ; then
 for cron_job_file in "${cron_job_dir}"/* ; do # */
 [[ -f "${cron_job_file}" ]] && echo "${cron_fields} ${cron_job_file}"
 done
 fi
 fi
 done;
 }
# Temporary file for crontab lines.
 temp=$(mktemp) || exit 1
# Add all of the jobs from the system-wide crontab file.
 cat "${CRONTAB}" | clean_cron_lines | lookup_run_parts >"${temp}"
# Add all of the jobs from the system-wide cron directory.
 cat "${CRONDIR}"/* | clean_cron_lines >>"${temp}" # */
# Add each user's crontab (if it exists). Insert the user's name between the
 # five time fields and the command.
 while read user ; do
 crontab -l -u "${user}" 2>/dev/null |
 clean_cron_lines |
 sed --regexp-extended "s/^((\S+ +){5})(.+)$/\1${user} \3/" >>"${temp}"
done <
# Output the collected crontab lines. Replace the single spaces between the
 # fields with tab characters, sort the lines by hour and minute, insert the
 # header line, and format the results as a table.
 cat "${temp}" |
 sed --regexp-extended "s/^(\S+) +(\S+) +(\S+) +(\S+) +(\S+) +(\S+) +(.*)$/\1\t\2\t\3\t\4\t\5\t\6\t\7/" |
 sort --numeric-sort --field-separator="${tab}" --key=2,1 |
 sed "1i\mi\th\td\tm\tw\tuser\tcommand" |
 column -s"${tab}" -t
rm --force "${temp}"

source: How do I list all cron jobs for all users?