How to convert time period string in sleep format to seconds in bash

Question

This is about a time string in the format accepted by the common linux command sleep, like "3d 7h 5m 10s" (3 days, 7 hours, 5 minutes and 10 seconds), which would have to result in:

(3 * 24 * 60 * 60) + (7 * 60 * 60) + (5 * 60) + 10 = 284710 seconds

Note that not all these 4 elements must be present, nor in the right order, and one element might appear multiple times. So "3s 5s 6h" is valid too, and should result in:

(6 * 60 * 60) + (3 + 5) = 21608 seconds

Answer 1

When you replace the letters with the corresponding factors, you can pipe that to bc. You only need to take care of the + at the end of the line.

t2s() {
   sed 's/d/*24*3600 +/g; s/h/*3600 +/g; s/m/*60 +/g; s/s/\+/g; s/+[ ]*$//g' <<< "$1" | bc
}

Testrun

$ t2s "3d 7h 5m 10s"
284710
$ t2s "3d 7h 5m "
284700
$ t2s "3s 5s 6h"
21608

Answer 2

I don't know if there is a predefined command that fits your requirements, but I came up with a more compact and loop-free script using GNU grep, paste, and bc.

Usage examples, assuming the script is saved as t2sec:

• t2sec 3s prints 3.
• t2sec "2m 1s", t2sec "1s 2m", t2sec "1m 1s 1m" all print 121.
• t2sec "1s 2s", t2sec "1s2s", t2sec 1s 2s, t2sec " 1s 2s " all print 3.
• t2sec, t2sec "", t2sec "1x", t2sec "s" all print nothing and exit with status of 1.

#! /bin/bash  

t="$*"

# validate
grep -Pqx '( *\d+[smhd])+ *' <<< "$t" || exit 1

# helper functions
sumAndMultiply() { bc <<< "(0$(paste -s -d+))*$1"; }
xToSeconds() { grep -Po "\\d+(?=$1)" | sumAndMultiply "$2"; }

# convert to seconds
(
        xToSeconds s 1 <<< "$t";
        xToSeconds m 60 <<< "$t";
        xToSeconds h 3600 <<< "$t";
        xToSeconds d 86400 <<< "$t";
) | sumAndMultiply 1