列出所有cronjobs，如果有任何沒有多餘的信息

Question

試圖輸出cronjobs列表，而不是用戶列表。 crontab -l的原始輸出太髒了，我似乎無法清理它。我運行這個sudo script.sh或su然後運行它。我也嘗試過調用它sudo script.sh | grep -v no。我很迷惑，爲什麼這不起作用：

#!/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 

BTW，有更簡單的方式來獲得代碼格式化正確這裏比在每行開頭的4位粘貼？我必須嘗試了40分鐘。不抱怨，只是問。

Answer 1

crontab將「沒有用戶的crontab」連接到標準錯誤。要擺脫這些消息，您可以在循環中運行

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

。

我也建議將USER重命名爲其他東西。 USER變量名稱是保留的，應該設置爲您的用戶（登錄）名稱。通常，您應該使用小寫變量名稱，以避免這種名稱衝突。