[SOLVED] Prase and report user login for today's date

bravored · 09-13-2013, 07:46 AM

All,

I want to get a count current user logon count for the day/date

The access.log doesnt rotate per day..

At the moment I get it by: cat access.log| grep 2013-09-13| grep Succeeded| wc -l

So for tomorrow I would have to change the date to 2013-09-14.

Any ideas?

Thanks,

joe_2000 · 09-13-2013, 08:44 AM

have you tried it with the date command? The manpage helps you to get the format right.

Habitual · 09-13-2013, 08:57 AM

http://www.linuxquestions.org/questi...-and-tar-41754
shows examples that you can work with.

bravored · 09-13-2013, 10:31 AM

Yea I tried the date command; I get syntax error

cat logon.list | grep 2013-09-11| |grep Succeeded | wc -l

OK I tried it again, and it went through.

Ended up using date=`date '+%Y-%m-%d'` as the variable

Habitual · 09-13-2013, 12:56 PM

Quote:

Originally Posted by bravored

Yea I tried the date command; I get syntax error

cat logon.list | grep 2013-09-11| |grep Succeeded | wc -l

OK I tried it again, and it went through.

Ended up using date=`date '+%Y-%m-%d'` as the variable

Don't cat, just to grep. Try

Code:

grep $(date +"%Y-%m-%d") logon.list | grep Succeeded | wc -l

Other more experienced people may have other ideas as well.