Convert time stamp to seconds in bash

As far as I know, bash variables do not have a certain type.

My problem is, I need to convert a time stamp (always in format: hh:mm:ss) to a single integer representing seconds.

So I cut the hh, mm and ss parts into separate strings and use expr to calculate the seconds integer like:

TIME=expr $HH \* 3600 + $MM \* 60 + $SS

But when e.g. $HH=00 then expr won't work. Is it possible to convert strings like 00 or 01 to integers?

Try awk. For example:

echo "00:20:40" | awk -F: '{ print ($1 * 3600) + ($2 * 60) + $3 }'

So, if you have:

time="00:20:40"

then:

seconds=$(echo $time | awk -F: '{ print ($1 * 3600) + ($2 * 60) + $3 }')

and echo $seconds will print 1240.

If you want to convert the current time into second, try this script

#!/bin/bash

let var=$(date +%H)*3600+$(date +%M)*60+$(date +%S)
echo $var

It will convert the current time into seconds (an integer).

If you wish to convert an arbitrary time string like HH:MM:SS into second then it is better to use Radu Rădeanu's answer. I could give something in the way you were trying,

I am assuming you stored hour, min, and second in HH, MM and SS, then use the following to store time in second into var,

let var=HH*3600+MM*60+SS

Convert time stamp to seconds in bash

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