I nee to subtract two lines which is in the format of time in shell. The time format is hh:mm:ss I used the code below to get the time.

cat /var/log/kern.log |grep usb |tail -2| awk '{print $3}'

The output of the above code is


How can I find the difference in seconds?


4 Answers 4


I would take a step further (inspired by this post):

# => 18:23:24 --> 66204
grep usb /var/log/kern.log|tail -2|awk '{print $3}'|awk -F: '{print ($1 * 3600) + ($2 * 60) + $3 }'

So, after I had:


You could then do:

echo $((66204-66020)) # => 184
  • 2
    Why are you piping awk into awk ?
    – user78605
    Nov 10, 2014 at 23:34
  • 1
    I'm separating concerns: the first awk gives me the hh:mm:ss format. The second awk transforms the hhh:mm:ss into integers. For me, it cuts down on cognitive load. Nov 11, 2014 at 16:59
  • @DarkHeart: what would you recommend? Nov 12, 2014 at 18:16

A more general solution because it works with times from different dates, too:

echo "18:23:24
18:20:20" | 
  (read later_time; read former_time;
    former_seconds=$(date --date="$former_time" +%s);
    later_seconds=$(date --date="$later_time" +%s);
    echo $((later_seconds-former_seconds)) )

If you use awk you can do it in one invocation

awk '
("date +%s -d "$3)|getline t[++i];
print t[i-1]-t[i];
}' /var/log/kern.log

Your entire pipeline in awk without lots of unnecessary pipes.
Similar to costas but without using any external commands.

awk '/usb/{split($3,a,":");b[++x]=(a[1]*3600)+(a[2]*60)+a[3]}END{print b[x-1]-b[x]}' /var/log/kern.log

Heres just the last bit with the dates worked out if you decide to keep the pipes

 awk '{split($1,a,":");b[x++]=(a[1]*3600)+(a[2]*60)+a[3]}END{print b[0]-b[1]}'
  • @Costas, there are only ever going to be two at the end of that pipe so it works fine, also the $2 was a typo and has been fixed :)
    – user78605
    Nov 11, 2014 at 10:30
  • Sorry I miss idea to remail all pipes
    – Costas
    Nov 11, 2014 at 11:03

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