列出所有 cronjobs（如果有没有多余信息）

Question

正在尝试输出 cronjobs 列表，而不是用户列表。 crontab -l 的原始输出太脏了，我似乎无法清理它。我运行使用 sudo script.sh 或 su 然后运行它。我也试过调用它 sudo script.sh | grep -v no。我很困惑为什么这不起作用：

#!/bin/bash
#Trying to show all cronjobs but no extraneous info
#
# This shows "no crontab for USER" for every USER without
# a crontab - I only want to see actual cronjobs, not a long
# list of users without crontabs
echo "Here is the basic output that needs manipulation:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
    crontab -l -u $USER
done
#
# grep -v fails me
# (grep'ing the output of the script as a whole fails also)
echo "
trying with grep -v no on each line:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
    crontab -l -u $USER | grep -v no
done

echo "
maybe with quotes around the no:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
    crontab -l -u $USER | grep -v "no"
done

# string manipulation - I can't even get started
echo "
And here I try to put the commmand output into a string so I can   manipulate it further, and use an if/then/fi on the product:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
    STRING="$(crontab -l -u $USER | grep -v no)"
    echo "STRING: $STRING"
done

顺便说一句，有没有比在每行开头粘贴 4 个空格更容易让代码在此处正确格式化的方法？我一定已经试验了 40 分钟。不抱怨，只求。

Answer 1

crontab 正在将 "no crontab for user" 连接到标准错误。要删除这些消息，您可以运行

crontab -l -u $USER 2>/dev/null

在你的循环中。

我还建议将 USER 重命名为其他名称。 USER 变量名是保留的，应设置为您的用户（登录）名。通常，您应该使用小写的变量名，以避免这种名称冲突。

列出所有 cronjobs（如果有没有多余信息）

List all cronjobs if there are any without superfluous info

bash

cron

grep

crontab

filter