It there a way to list jobs that has been
executed by cron or anacron in CentOS7 for
a given period of time, e.g. last 24 hours?

or equivalent.

Solved!

This has been solved but if someone would came across with same issue try something like this:

2 members found this post helpful.

Good idea, but what if cron wasn't executed at that particular hour of the day?

How do you use awk to do a lines range
search instead of sed for above?

I guess you can adjust the time range, but more specific time will get specific result. Log file can be very big, if the output is quite long then it's quite difficult to see or check.

Try something like this:

awk '/(Apr 10 00:00:01)/ { print $0 }' /var/log/cron

Please see links below:
http://quickbytesstuff.blogspot.sg/2...-tutorial.html

http://quickbytesstuff.blogspot.sg/2...e-example.html

That's not the point. The problem is that if there are no logs at "/Apr 10 00:00:01/", then absolutely nothing will show up. Moreover, if there are no logs at 'Apr 11 00:01:01', then all lines will show up until there are no more lines in the log. Not really nice.

I had been looking for the same answer for a while. I guess one solution would be to simply do grep for the day you're interested and then pipe it into awk:
grep "Apr 10" | awk -F '[ :]' '$3 >= 15 && $4 >= 30 { print $0 }' - which is going to print all logs starting from 3:30 PM on April 10, even if there is are no logs at exactly 3:30.


As far as your second solution is concerned "awk '/(Apr 10 00:00:01)/ { print $0 }' /var/log/cron)", I really don't understand what you're trying to do. That will only print the lines containing "Apr 10 00:00:01". I don't see the point. He was asking about a line range.