Displaying seconds as days/hours/mins/seconds?
You can use something like this:
function displaytime {
local T=$1
local D=$((T/60/60/24))
local H=$((T/60/60%24))
local M=$((T/60%60))
local S=$((T%60))
(( $D > 0 )) && printf '%d days ' $D
(( $H > 0 )) && printf '%d hours ' $H
(( $M > 0 )) && printf '%d minutes ' $M
(( $D > 0 || $H > 0 || $M > 0 )) && printf 'and '
printf '%d seconds\n' $S
}
Examples:
$ displaytime 11617
3 hours 13 minutes and 37 seconds
$ displaytime 42
42 seconds
$ displaytime 666
11 minutes and 6 seconds
Easiest and cleanest way is this one liner (here assuming GNU date
):
If the number of seconds is, say:
seconds=123456789 # as in one of the answers above
eval "echo $(date -ud "@$seconds" +'$((%s/3600/24)) days %H hours %M minutes %S seconds')"
--> output: 1428 days 21 hours 33 minutes 09 seconds
Credit goes to Stéphane Gimenez but if someone would like to display seconds only if a period is less than a minute here is my modified version that I use (also with fixed pluralization):
converts()
{
local t=$1
local d=$((t/60/60/24))
local h=$((t/60/60%24))
local m=$((t/60%60))
local s=$((t%60))
if [[ $d > 0 ]]; then
[[ $d = 1 ]] && echo -n "$d day " || echo -n "$d days "
fi
if [[ $h > 0 ]]; then
[[ $h = 1 ]] && echo -n "$h hour " || echo -n "$h hours "
fi
if [[ $m > 0 ]]; then
[[ $m = 1 ]] && echo -n "$m minute " || echo -n "$m minutes "
fi
if [[ $d = 0 && $h = 0 && $m = 0 ]]; then
[[ $s = 1 ]] && echo -n "$s second" || echo -n "$s seconds"
fi
echo
}
An alternative example in POSIX:
converts(){
t=$1
d=$((t/60/60/24))
h=$((t/60/60%24))
m=$((t/60%60))
s=$((t%60))
if [ $d -gt 0 ]; then
[ $d = 1 ] && printf "%d day " $d || printf "%d days " $d
fi
if [ $h -gt 0 ]; then
[ $h = 1 ] && printf "%d hour " $h || printf "%d hours " $h
fi
if [ $m -gt 0 ]; then
[ $m = 1 ] && printf "%d minute " $m || printf "%d minutes " $m
fi
if [ $d = 0 ] && [ $h = 0 ] && [ $m = 0 ]; then
[ $s = 1 ] && printf "%d second" $s || printf "%d seconds" $s
fi
printf '\n'
}