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marinm · #1 (**permalink**) 03-22-2005, 06:05 PM

Hi everyone,

I have a small issue. I have a lot of users on my server and I would like to check out their crontab, but doing it manually would take a lifetime.
Can anyone help me with a small script which will read /etc/passwd and output every user's crontab? The command is crontab -u user -l I think.
Any ideas?

Thank you.

nixcraft · #2 (**permalink**) 03-23-2005, 11:52 AM

That is quite easy

Code:

cat /etc/passwd | cut -d: -f1 | xargs crontab -l -u

Test it and lemme know...

marinm · #3 (**permalink**) 03-23-2005, 02:35 PM

Hi,

I get an error when trying to execute the command:

root@orbit [~]# cat /etc/passwd | cut -d: -f1 | xargs crontab -l -u
crontab: usage error: no arguments permitted after this option

Kind Regards

nixcraft · #4 (**permalink**) 03-23-2005, 08:09 PM

Sorry try as follows

Code:

a=$(cat /etc/passwd | cut -d: -f1)
for i in $a
do
 crontab -u $i -l
done

This one should work, let me know...

brew · #5 (**permalink**) 07-19-2005, 08:20 AM

Quote:

Originally Posted by nixcraft

That is quite easy

Code:

cat /etc/passwd | cut -d: -f1 | xargs crontab -l -u

Test it and lemme know...

i try above in freebsd..this is what i got:

-bash-2.05b# cat /etc/passwd | cut -d: -fl | xargs crontab -l -u
cut: [-cf] list: illegal list value
-bash-2.05b#

permalink · #6 (**permalink**) 07-20-2005, 12:24 AM

Following is correct command for FreeBSD:

Code:

cut -d: -f1 /etc/passwd | grep -vE "#" | xargs -I {} crontab -u {} -l

And for linux it should be as follows:

Code:

cut -d: -f1 /etc/passwd | grep -vE "#" | xargs -i{} crontab -u {} -l

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list every user's crontab script

list every user's crontab script

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