Format output

[SAM_24]

Hi,

I want to remove the time fields from crontab -l output. For example

* * * * * command

I want to remove everything before command.

In place of * there'd be time specs. In command field there'd be different script name.

Any help is appreciated.

Thanks.

[Sundar_7]

May not be the most elegant way, but gets the job done :-)

crontab -l | egrep -v "^#|^$" | awk '{for(i=6;i<=NF;i++) printf("%s ",$i); printf("\n");}'

[BONNAFOUS Jean Marc]

Hi,

crontab -l|egrep -v "^#" |cut -d" " -f6


Best Regards
JMB

[Franky_1]

Hi,

just type

crontab -l|grep -v "^#"|awk '{print $6}'

Should do the trick

Regards

Franky

[Muthukumar_5]

We can do this as,
crontab -l | grep "^[ 0-9*]" | cut -d" " -f6-

It will cut all field's 6th and after

-------
Franky, marc, we will be having some more filed's after 6 there on cron too :-)

05,15,25,35,45,55 * * * * /usr/sbin/dmesg - >>/var/adm/messages

will give only /usr/sbin/dmesg in your try there.
---------

HTH.
Muthu

[Rodney Hills]

How about a perl one-liner:

crontab -l | perl -ne 'split("\\s+",$_,6); print $_[5]'

HTH

-- Rod Hills