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vnix27
Asked: 2011-06-02 03:05:16 +0800 CST

Calculate highest time difference between log message in /var/log/messages

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I need to calculate time difference between two log messages of /var/log/messages. As we know log messages prefixed by date & time , i want time difference between two log entries.

linux command

3 Answers

  1. In bash, once you figure out what date/times you want to use you can write

    firsttime="Jun  1 02:00:13"
secondtime="Jun  1 03:00:46"
echo $(( $(date -d "$secondtime" +"%s") - $(date -d "$firsttime" +"%s") ))

    Which will give you number of seconds between those two times. If you want to automate finding firsttime and secondtime, you're going to have to tell us what exactly you're looking for in the log.

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  2. I assume you want to find the largest time gap between two consecutive log messages. If so, see below for a script that should do the trick. I wrote this script a couple of years ago to print out a particular time range from the messages file, and it was pretty easy to customize:

    #!/usr/bin/perl -W

use strict;
use Time::Local;

my %MONTHS = ( "Jan" => 0, "Feb" => 1, "Mar" => 2, "Apr" => 3, "May" => 4, "Jun" => 5,
               "Jul" => 6, "Aug" => 7, "Sep" => 8, "Oct" => 9, "Nov" => 10, "Dec" => 11 );

# Messages file doesn't include the year, so we need to assume that
# all messages are less than a year old and make some guesses

my $currenttime = time();
my $currentyear  = (localtime($currenttime))[5];
my $currentmonth = (localtime($currenttime))[4];

my $largestgap;
my $largestgapbefore;
my $largestgapafter;
my $lasttime;
my $lastmsg;

open(FILE, "</var/log/messages") || die "Unable to open messages file: $!\n";
while(<FILE>) {
    chomp;
    /^(...) (..) (..):(..):(..)/;
    my $year = $currentyear;
    if($MONTHS{$1} > $currentmonth) {$year -= 1; } 
    my $time = timelocal($5, $4, $3, $2, $MONTHS{$1}, $year);

    if(defined($lasttime)) {
        my $gap = $time - $lasttime;
        if(!defined($largestgap) || $gap > $largestgap) {
            $largestgap = $gap;
            $largestgapbefore = $lastmsg;
            $largestgapafter  = $_;
        }
    }
    $lasttime = $time;
    $lastmsg  = $_;
}
close(FILE);

if(!defined($lasttime)) {
    print "No entries in log file.\n";
    exit;
}

print "Largest gap was: " . $largestgap . " seconds.\n";

print "Entry before the gap: " . $largestgapbefore . "\n";
print " Entry after the gap: " . $largestgapafter . "\n";
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  3. Best Answer

    I've chosen the first and last messages in /var/log/messages to demonstrate, you'll have to modify it to get your own start and end times

    #!/bin/bash

start=$(head -n 1 /var/log/messages | awk {'print $1, $2,$3'})
finish=$(tail -n 1 /var/log/messages | awk {'print $1, $2,$3'})

start_secs=$(date -d "$start" +"%s")
finish_secs=$(date -d  "$finish" +"%s")

let diff_secs=($finish_secs - $start_secs)

let days=$diff_secs/86400
let remainder=$diff_secs%86400
let hours=$remainder/3600
let remainder=$remainder%3600
let minutes=$remainder/60
let seconds=$remainder%60
echo "Days = $days, Hours = $hours, Minutes = $minutes, Seconds = $seconds"
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