How to subtract two times in shell

Question

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

18:23:24
18:20:20

How can I find the difference in seconds?

Solution 1

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:

66204
66020

You could then do:

echo $((66204-66020)) # => 184

Solution 2

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)) )
184

Comments

[Admin]

Why are you piping awk into awk ?

[Admin]

@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 :)

[Costas]

Sorry I miss idea to remail all pipes

[Soinou]

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.

[Soinou]

@DarkHeart: what would you recommend?