If you need a list of all users crontab tasks, run this as root:
for user in $(cut -f1 -d: /etc/passwd); do crontab -u $user -l; done
will loop over each user name listing out their crontab. The crontabs are owned by the respective users so you won’t be able to see another user’s crontab w/o being them or root.
–[edit] if you want to know, which user does a crontab belong to insert echo $user
for user in $(cut -f1 -d: /etc/passwd); do echo $user; crontab -u $user -l; done
As a side note, this doesn’t work when the users are defined in NIS or LDAP. You need to use
for user in $(getent passwd | cut -f1 -d: ); do echo $user; crontab -u $user -l; done
I have also seen this bash script that supposedly takes into account displaying other crons, (including the scripts launched by run-parts in /etc/cron.hourly, /etc/cron.daily, etc.) and the jobs in the /etc/cron.d.
#!/bin/bash # System-wide crontab file and cron job directory. Change these for your system. CRONTAB='/etc/crontab' CRONDIR='/etc/cron.d' # Single tab character. Annoyingly necessary. tab=$(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 "${cron_job_dir}"/* ; do # */ <not a comment> [[ -f "${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}" # */ <not a comment> # 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 < <(cut --fields=1 --delimiter=: /etc/passwd) # 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}"
I have not thoroughly tested this script, I ran it once on my server, I found it here.
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