我喜欢上面简单的一行字回答:

$(cut -f1 -d: /etc/passwd);执行crontab -u $user -l;完成

但是Solaris没有-u标志，也不会打印正在检查的用户，你可以这样修改:

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

当一个帐户不允许使用cron等时，您将得到一个用户列表，其中没有crontab抛出的错误。注意，在Solaris中，角色也可以在/etc/passwd中(参见/etc/user_attr)。

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

请注意，它显示了用户，并或多或少地按小时和分钟排序，以便我可以看到每天的日程安排。

到目前为止，我已经在Ubuntu、Debian和Red Hat AS上测试了它。

#!/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}"

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

这避免了直接与passwd混淆，跳过没有cron条目的用户，对于那些有cron条目的用户，它会打印出用户名以及crontab。

主要是把这个放在这里，虽然这样我可以找到它，以防我需要再次搜索它。

下面剥离了不使用crontab的用户的注释、空行和错误。剩下的就是一个清晰的用户列表和他们的工作。

注意，在第二行中使用了sudo。如果你已经是根用户了，移除它。

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

示例输出:

# ------ 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

我在Ubuntu(12到16)和Red Hat(5到7)上使用这个。

这将显示所有用户的所有crontab条目。

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

在Solaris上，对于特定的已知用户名:

crontab -l username

要在Solaris上一次性获得所有用户的作业，就像上面的其他文章一样:

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

更新: 请停止建议Solaris上的错误编辑: