Bash - convert time interval string to nr. of seconds - bash

Question

I'm trying to convert strings, describing a time interval, to the corresponding number of seconds.
After some experimenting I figured out that I can use date like this:
soon=$(date -d '5 minutes 10 seconds' +%s); now=$(date +%s)
echo $(( $soon-$now ))
but I think there should be an easier way to convert strings like "5 minutes 10 seconds" to the corresponding number of seconds, in this example 310. Is there a way to do this in one command?
Note: although portability would be useful, it isn't my top priority.

Answer 1

You could start at epoch
date -d"1970-01-01 00:00:00 UTC 5 minutes 10 seconds" "+%s"
310
You could also easily sub in times
Time="1 day"
date -d"1970-01-01 00:00:00 UTC $Time" "+%s"
86400

Answer 2

There is one way to do it, without using date command in pure bash (for portability)
Assuming you just have an input string to convert "5 minutes 10 seconds" in a bash variable with a : de-limiter as below.
$ convertString="00:05:10"
$ IFS=: read -r hour minute second <<< "$convertString"
$ secondsValue=$(((hour * 60 + minute) * 60 + second))
$ printf "%s\n" "$secondsValue"
310
You can run the above commands directly on the command-line without the $ mark.

Answer 3

This will do (add the epoch 19700101):
$ date -ud '19700101 5 minutes 10 seconds' +%s
310
It is important to add a -u to avoid local time (and DST) effects.
$ TZ=America/Los_Angeles date -d '19700101 5 minutes 10 seconds' +%s
29110
Note that date could do some math:
$ date -ud '19700101 +5 minutes 10 seconds -47 seconds -1 min' +%s
203

Answer 4

The previous suggestions didn't work properly on alpine linux, so here's a small helper function that is POSIX compliant, is easy to use and also supports calculations (just as a side effect of the implementation).
The function always returns an integer based on the provided parameters.
$ durationToSeconds '<value>' '<fallback>'
$ durationToSeconds "1h 30m"
5400
$ durationToSeconds "$someemptyvar" 1h
3600
$ durationToSeconds "$someemptyvar" "1h 30m"
5400
# Calculations also work
$ durationToSeconds "1h * 3"
10800
$ durationToSeconds "1h - 1h"
0
# And also supports long forms for year, day, hour, minute, second
$ durationToSeconds "3 days 1 hour"
262800
# It's also case insensitive
$ durationToSeconds "3 Days"
259200
function durationToSeconds () {
set -f
normalize () { echo $1 | tr '[:upper:]' '[:lower:]' | tr -d "\"\\\'" | sed 's/years\{0,1\}/y/g; s/months\{0,1\}/m/g; s/days\{0,1\}/d/g; s/hours\{0,1\}/h/g; s/minutes\{0,1\}/m/g; s/min/m/g; s/seconds\{0,1\}/s/g; s/sec/s/g; s/ //g;'; }
local value=$(normalize "$1")
local fallback=$(normalize "$2")
echo $value | grep -v '^[-+*/0-9ydhms]\{0,30\}$' > /dev/null 2>&1
if [ $? -eq 0 ]
then
>&2 echo Invalid duration pattern \"$value\"
else
if [ "$value" = "" ]; then
[ "$fallback" != "" ] && durationToSeconds "$fallback"
else
sedtmpl () { echo "s/\([0-9]\+\)$1/(0\1 * $2)/g;"; }
local template="$(sedtmpl '\( \|$\)' 1) $(sedtmpl y '365 * 86400') $(sedtmpl d 86400) $(sedtmpl h 3600) $(sedtmpl m 60) $(sedtmpl s 1) s/) *(/) + (/g;"
echo $value | sed "$template" | bc
fi
fi
set +f
}

Answer 5

Edit : Yes. I developed for OP after comment and checked on Mac OS X, CentOS and Ubuntu. One liner, POSIX compliant command for converting "X minutes Y seconds" format to seconds. That was the question.
echo $(($(echo "5 minutes 10 seconds" | cut -c1-2)*60 + $(echo "5 minutes 10 seconds" | cut -c1-12 | awk '{print substr($0,11)}')))
OP told me via comment that he wants for "X minutes Y seconds" format not for HH:MM:SS format. The command with date and "+%s" is throwing error on (my) Mac. OP wanted to grab the numerical values from "X minutes Y seconds" format and convert it to seconds. First I extracted the minute in digit (take it as equation A) :
echo "5 minutes 10 seconds" | cut -c1-2)
then I extracted the seconds part (take it as equation B) :
echo "5 minutes 10 seconds" | cut -c1-12 | awk '{print substr($0,11)}'
Now multiply minute by 60 then add with the other :
echo $((equation A)*60) + (equation B))
OP should ask the others to check my developmental version (but working) of command before using it for automatic repeated usage like we do with cron on a production server.
If we want to run this on a log file with values in "X minutes Y seconds" format, we have to change echo "5 minutes 10 seconds" to cat file | ... like command. I kept a gist of it too if I or others ever need we can use it with cat to run on server log files with x minutes y seconds like log format.
Although off-topic (what I understood, question has not much to do with current time), this is not working for POSIX-compliant OS to get current time in seconds :
date -d "1970-01-01 00:00:00 UTC 5 minutes 10 seconds" "+%s"
It will throw error on MacOS X but work on most GNU/Linux distro. That +%s part will throw error on POSIX-compliant OS upon complicated usage. These commands are mostly suitable to get current time in seconds on POSIX compliant to any kind of unix like OS :
awk 'BEGIN{srand(); print srand()}'
perl -le 'print time'
If OP needs can extend it by generating current time in seconds and subtract. I hope it will help.
---- OLD Answer before EDIT ----
You can get the current time without that date -- echo | awk '{print systime();}' or wget -qO- http://www.timeapi.org/utc/now?\\s. Other way to convert time to second is echo "00:20:40.25" | awk -F: '{ print ($1 * 3600) + ($2 * 60) + $3 }'.
The example with printf shown in another answer is near perfect.
That thing you want is always needed by the basic utilities of GNU/Linux - gnu.org/../../../../../Setting-an-Alarm.html
Way to approach really depends how much foolproof way you need.