How do I list all cron jobs for all users?

Question

Is there a command or an existing script that will let me view all of a *NIX system's scheduled cron jobs at once? I'd like it to include all of the user crontabs, as well as /etc/crontab, and whatever's in /etc/cron.d. It would also be nice to see the specific commands run by run-parts in /etc/crontab.

Ideally, I'd like the output in a nice column form and ordered in some meaningful way.

I could then merge these listings from multiple servers to view the overall "schedule of events."

I was about to write such a script myself, but if someone's already gone to the trouble...

Answer 1

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:

mi     h    d  m  w  user      command
09,39  *    *  *  *  root      [ -d /var/lib/php5 ] && find /var/lib/php5/ -type f -cmin +$(/usr/lib/php5/maxlifetime) -print0 | xargs -r -0 rm
47     */8  *  *  *  root      rsync -axE --delete --ignore-errors / /mirror/ >/dev/null
17     1    *  *  *  root      /etc/cron.daily/apt
17     1    *  *  *  root      /etc/cron.daily/aptitude
17     1    *  *  *  root      /etc/cron.daily/find
17     1    *  *  *  root      /etc/cron.daily/logrotate
17     1    *  *  *  root      /etc/cron.daily/man-db
17     1    *  *  *  root      /etc/cron.daily/ntp
17     1    *  *  *  root      /etc/cron.daily/standard
17     1    *  *  *  root      /etc/cron.daily/sysklogd
27     2    *  *  7  root      /etc/cron.weekly/man-db
27     2    *  *  7  root      /etc/cron.weekly/sysklogd
13     3    *  *  *  archiver  /usr/local/bin/offsite-backup 2>&1
32     3    1  *  *  root      /etc/cron.monthly/standard
36     4    *  *  *  yukon     /home/yukon/bin/do-daily-stuff
5      5    *  *  *  archiver  /usr/local/bin/update-logs >/dev/null

Note that it shows the user, and more-or-less sorts by hour and minute so that I can see the daily schedule.

So far, I've tested it on Ubuntu, Debian, and Red Hat AS.

#!/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}"

Answer 2

You would have to run this as root, but:

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 a crontab belongs to, use echo $user

for user in $(cut -f1 -d: /etc/passwd); do echo $user; crontab -u $user -l; done

Answer 3

Under Ubuntu or debian, you can view crontab by /var/spool/cron/crontabs/ and then a file for each user is in there. That's only for user-specific crontab's of course.

For Redhat 6/7 and Centos, the crontab is under /var/spool/cron/.

Answer 4

This will show all crontab entries from all users.

sed 's/^\([^:]*\):.*$/crontab -u \1 -l 2>\&1/' /etc/passwd | grep -v "no crontab for" | sh

Answer 5

Depends on your linux version but I use:

tail -n 1000 /var/spool/cron/*

as root. Very simple and very short.

Gives me output like:

==> /var/spool/cron/root <==
15 2 * * * /bla

==> /var/spool/cron/my_user <==
*/10 1 * * * /path/to/script

Answer 6

A small refinement of Kyle Burton's answer with improved output formatting:

#!/bin/bash
for user in $(cut -f1 -d: /etc/passwd)
do echo $user && crontab -u $user -l
echo " "
done

Answer 7

getent passwd | cut -d: -f1 | perl -e'while(<>){chomp;$l = `crontab -u $_ -l 2>/dev/null`;print "$_\n$l\n" if $l}'

This avoids messing with passwd directly, skips users that have no cron entries and for those who have them it prints out the username as well as their crontab.

Mostly dropping this here though so i can find it later in case i ever need to search for it again.

Answer 8

If you check a cluster using NIS, the only way to see if a user has a crontab entry ist according to Matt's answer /var/spool/cron/tabs.

grep -v "#" -R  /var/spool/cron/tabs

Answer 9

This script worked for me in CentOS to list all crons in the environment:

sudo cat /etc/passwd | sed 's/^\([^:]*\):.*$/sudo crontab -u \1 -l 2>\&1/' | grep -v "no crontab for" | sh

Answer 10

I like the simple one-liner answer above:

for user in $(cut -f1 -d: /etc/passwd); do crontab -u $user -l; done

But Solaris which does not have the -u flag and does not print the user it's checking, you can modify it like so:

for user in $(cut -f1 -d: /etc/passwd); do echo User:$user; crontab -l $user 2>&1 | grep -v crontab; done

You will get a list of users without the errors thrown by crontab when an account is not allowed to use cron etc. Be aware that in Solaris, roles can be in /etc/passwd too (see /etc/user_attr).

Answer 11

for user in $(cut -f1 -d: /etc/passwd); 
do 
    echo $user; crontab -u $user -l; 
done

Answer 12

To get list from ROOT user.

for user in $(cut -f1 -d: /etc/passwd); do echo $user; sudo crontab -u $user -l; done

Answer 13

The following strips away comments, empty lines, and errors from users with no crontab. All you're left with is a clear list of users and their jobs.

Note the use of sudo in the 2nd line. If you're already root, remove that.

for USER in $(cut -f1 -d: /etc/passwd); do \
USERTAB="$(sudo crontab -u "$USER" -l 2>&1)";  \
FILTERED="$(echo "$USERTAB"| grep -vE '^#|^$|no crontab for|cannot use this program')";  \
if ! test -z "$FILTERED"; then  \
echo "# ------ $(tput bold)$USER$(tput sgr0) ------";  \
echo "$FILTERED";  \
echo "";  \
fi;  \
done

Example output:

# ------ root ------
0 */6 * * * /usr/local/bin/disk-space-notify.sh
45 3 * * * /opt/mysql-backups/mysql-backups.sh
5 7 * * * /usr/local/bin/certbot-auto renew --quiet --no-self-upgrade

# ------ sammy ------
55 * * * * wget -O - -q -t 1 https://www.example.com/cron.php > /dev/null

I use this on Ubuntu (12 thru 16) and Red Hat (5 thru 7).

Answer 14

Depends on your version of cron. Using Vixie cron on FreeBSD, I can do something like this:

(cd /var/cron/tabs && grep -vH ^# *) 

if I want it more tab deliminated, I might do something like this:

(cd /var/cron/tabs && grep -vH ^# * | sed "s/:/      /")

Where that's a literal tab in the sed replacement portion.

It may be more system independent to loop through the users in /etc/passwd and do crontab -l -u $user for each of them.

Answer 15

Thanks for this very useful script. I had some tiny problems running it on old systems (Red Hat Enterprise 3, which handle differently egrep and tabs in strings), and other systems with nothing in /etc/cron.d/ (the script then ended with an error). So here is a patch to make it work in such cases :

2a3,4
> #See:  http://stackoverflow.com/questions/134906/how-do-i-list-all-cron-jobs-for-all-users
>
27c29,30
<         match=$(echo "${line}" | egrep -o 'run-parts (-{1,2}\S+ )*\S+')
---
>         #match=$(echo "${line}" | egrep -o 'run-parts (-{1,2}\S+ )*\S+')
>         match=$(echo "${line}" | egrep -o 'run-parts.*')
51c54,57
< cat "${CRONDIR}"/* | clean_cron_lines >>"${temp}"  # */ <not a comment>
---
> sys_cron_num=$(ls /etc/cron.d | wc -l | awk '{print $1}')
> if [ "$sys_cron_num" != 0 ]; then
>       cat "${CRONDIR}"/* | clean_cron_lines >>"${temp}"  # */ <not a comment>
> fi
67c73
<     sed "1i\mi\th\td\tm\tw\tuser\tcommand" |
---
>     sed "1i\mi${tab}h${tab}d${tab}m${tab}w${tab}user${tab}command" |

I'm not really sure the changes in the first egrep are a good idea, but well, this script has been tested on RHEL3,4,5 and Debian5 without any problem. Hope this helps!

Answer 16

Building on top of @Kyle

for user in $(tail -n +11 /etc/passwd | cut -f1 -d:); do echo $user; crontab -u $user -l; done

to avoid the comments usually at the top of /etc/passwd,

And on macosx

for user in $(dscl . -list /users | cut -f1 -d:); do echo $user; crontab -u $user -l; done    

Answer 17

I think a better one liner would be below. For example if you have users in NIS or LDAP they wouldnt be in /etc/passwd. This will give you the crontabs of every user that has logged in.

for I in `lastlog | grep -v Never | cut -f1 -d' '`; do echo $I ; crontab -l -u $I ; done

Answer 18

you can write for all user list :

sudo crontab -u userName -l

,

You can also go to

cd /etc/cron.daily/
ls -l
cat filename

this file will list the schedules

cd /etc/cron.d/
ls -l
cat filename

Answer 19

With apologies and thanks to yukondude.

I've tried to summarise the timing settings for easy reading, though it's not a perfect job, and I don't touch 'every Friday' or 'only on Mondays' stuff.

This is version 10 - it now:

• runs much much faster
• has optional progress characters so you could improve the speed further.
• uses a divider line to separate header and output.
• outputs in a compact format when all timing intervals uencountered can be summarised.
• Accepts Jan...Dec descriptors for months-of-the-year
• Accepts Mon...Sun descriptors for days-of-the-week
• tries to handle debian-style dummying-up of anacron when it is missing
• tries to deal with crontab lines which run a file after pre-testing executability using "[ -x ... ]"
• tries to deal with crontab lines which run a file after pre-testing executability using "command -v"
• allows the use of interval spans and lists.
• supports run-parts usage in user-specific /var/spool crontab files.

I am now publishing the script in full here.

https://gist.github.com/myshkin-uk/d667116d3e2d689f23f18f6cd3c71107

Answer 20

Since it is a matter of looping through a file (/etc/passwd) and performing an action, I am missing the proper approach on How can I read a file (data stream, variable) line-by-line (and/or field-by-field)?:

while IFS=":" read -r user _
do
   echo "crontab for user ${user}:"
   crontab -u "$user" -l
done < /etc/passwd

This reads /etc/passwd line by line using : as field delimiter. By saying read -r user _, we make $user hold the first field and _ the rest (it is just a junk variable to ignore fields).

This way, we can then call crontab -u using the variable $user, which we quote for safety (what if it contains spaces? It is unlikely in such file, but you can never know).

Answer 21

On Solaris, for a particular known user name:

crontab -l username

All other *Nix will need -u modifier:

crontab -u username -l

To get all user's jobs at once on Solaris, much like other posts above:

for user in $(cut -f1 -d: /etc/passwd); do crontab -l $user 2>/dev/null; done

Answer 22

For me look at /var/spool/cron/crontabs is the best way

Answer 23

This script outputs the Crontab to a file and also lists all users confirming those which have no crontab entry:

for user in $(cut -f1 -d: /etc/passwd); do 
  echo $user >> crontab.bak
  echo "" >> crontab.bak
  crontab -u $user -l >> crontab.bak 2>> > crontab.bak
done