Shell Programming and Scripting

Hi all, I have tried to figure out a way to automatically get the last 5 minutes' log content from log file, at first, my thoughts like this,

sed -n "/ $(date +\%R -d "-5 min")/,$"p syslog > newfile,

but quickly I found it did not work, say I have a syslog file as following,

Jul 19 18:15:23 ........
Jul 19 18:15:28 ........
Jul 19 18:17:11 ........
Jul 19 18:18:11 ........

if the script run at 18:21，18:21 minus 5minutes is 18:16，since there is no "18:16" in logfile, sed will find no match, thus there will be nothing in the newfile", although actually it should have!

I know there will be solution around the corner, but how? Thanks!

Well since there are only 5 minutes you are interested in:

in bash

Code:

for (( i = 5; i >=0; i-- )) ; do
     grep $(date +%R -d "-$i  min") syslog >> newfile
done

Hi reborg, thank you for your reply,

I am just wondering is there any other better solution other then using "grep",

If the syslog file is super large, doing loop and grep maybe not a very good solution.

Try this one

Code:

#!/bin/sh

e=
for (( i = 5; i >= 0; i-- )) ; do
    e='-e /'`date +\%R -d "-$i min"`'/p '$e
done

$(sed -n $e syslog > newfile)

This script generates commands like

Code:

sed -n -e /23:40/p -e /23:39/p -e /23:38/p -e /23:37/p -e /23:36/p -e /23:35/p  syslog > newfile

Well you could use egrep and multiple patterns to only search once.

assign each of the last 5 minutes to a variable and do

egrep "var1|var2|var3|var4|var5" syslog > newlog

or just take the last line of the previous extracted log.

Code:

var=$(tail -1 last_log_file)
sed -ne "/$var/"',$p' syslog > newlog

AH! much better, thanks, you guys!