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# 1  
Old 03-05-2013
Part of log
hi,

i have a requirement where i need to grep the part of the log file of last 5 minutes back.

Code:
[10/2/08 9:40:39:865 EDT] 00000011 ViewReceiver  I   DCSV1033I: DCS Stack DefaultCoreGroup at Member usa0300ux423Cell01\usa0300ux423QASrv02\server1: Confirmed all new view members in view identifier (11:0.usa0300ux423Cell01\usa0300ux423QASrv01\server1). View channel type is View|Ptp.

please suggest??
Last edited by Scrutinizer; 03-05-2013 at 08:18 AM.. Reason: code tags
# 2  
Old 03-05-2013
Code:
awk '$0>=from && $0<=to' from="$(date +"[%x %H:%M:%S" -d -5min)" to="$(date +"[%x %H:%M:%S")" logfile

You may need to tune the date format to reflect your log
Last edited by Jotne; 03-05-2013 at 09:10 AM..
# 3  
Old 03-05-2013
thanks for the reply Jotne, but i have 1 question here:
Code:
%x %H:%M:%S" -d 5min

in the from clause you have written this code but it only gives the current time, how
Code:
bash-3.2$ date +"[%x %H:%M:%S" -d 5min
[03/05/13 08:00:06
bash-3.2$ date
Tue Mar  5 08:00:13 EST 2013

how could i go back 5 minutes?
Last edited by Franklin52; 03-05-2013 at 09:41 AM.. Reason: Please use code tags for data and code samples
# 4  
Old 03-05-2013
Use Code tag please
My line above take the last 5 minutes form the log -d -5min
It goes form now and 5 minute back and return every line within that time frame.

This give now date +"[%x %H:%M:%S" [03/05/13 14:06:10
This give -5 min date +"[%x %H:%M:%S" -d -5min [03/05/13 14:01:10

Edit
Fixed small typo, missing a -
# 5  
Old 03-05-2013
but i am getting the same result as i was getting earlier:
Code:
bash-3.2$ date +"[%x %H:%M:%S" -d -5min
[03/05/13 09:20:51
bash-3.2$ date
Tue Mar  5 09:21:00 EST 2013

may be because i am working on solaris env??
please help with some more ideas..
Last edited by Franklin52; 03-05-2013 at 10:49 AM.. Reason: Please use code tags for data and code samples
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