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Old 04-11-2017, 12:38 AM   #1
fanoflq
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CentOS 7: How to list completed cron jobs?


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?
 
Old 04-11-2017, 12:50 AM   #2
descendant_command
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Code:
# grep CRON /var/log/syslog
or equivalent.
 
Old 04-11-2017, 01:20 AM   #3
fanoflq
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Solved!
Quote:
/var/log/cron
 
Old 04-11-2017, 03:10 AM   #4
JJJCR
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This has been solved but if someone would came across with same issue try something like this:

Quote:
sed -n '/Apr 10 00:00:01/ , /Apr 11 00:01:01/p' /var/log/cron

Last edited by JJJCR; 04-11-2017 at 03:10 AM. Reason: edit
 
2 members found this post helpful.
Old 04-11-2017, 03:57 AM   #5
vincix
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Quote:
Originally Posted by JJJCR View Post
This has been solved but if someone would came across with same issue try something like this:
Good idea, but what if cron wasn't executed at that particular hour of the day?
 
Old 04-11-2017, 11:38 AM   #6
fanoflq
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Quote:
Originally Posted by JJJCR View Post
This has been solved but if someone would came across with same issue try something like this:
How do you use awk to do a lines range
search instead of sed for above?
 
Old 04-17-2017, 01:33 AM   #7
JJJCR
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Cool

Quote:
Originally Posted by vincix View Post
Good idea, but what if cron wasn't executed at that particular hour of the day?
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.
 
Old 04-17-2017, 03:19 AM   #8
JJJCR
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Smile

Quote:
Originally Posted by fanoflq View Post
How do you use awk to do a lines range
search instead of sed for above?
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
 
Old 04-17-2017, 01:17 PM   #9
vincix
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Quote:
Originally Posted by JJJCR View Post
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.
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.
 
  


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