Korn Shell: Show elapsed time in a specific format

Ksh has a special parameter SECONDS which always contains the number of seconds since the epoch. Evaluating $SECONDS at the beginning and at the end of the job gives you the start and end times, and the difference is the elapsed time.

Unix time doesn't take leap seconds into account: a day in Unix time is always exactly 86400 seconds. Therefore time-of-day arithmetic on Unix time is easy.

start=$SECONDS
…
end=$SECONDS
elapsed=$((end - start))
printf 'Process completed %d:%02d:%02d - Elapsed %d:%02d:%02d\n' \
       $((end / 3600)) $((end / 60 % 60)) $((end % 60)) \
       $((elapsed / 3600)) $((elapsed / 60 % 60)) $((elapsed % 60))

Like this (Warning: those of a sensitive disposition look away now - this is old school ksh):

t1=`date '+%H:%M:%S'`
sleep 2
t2=`date '+%H:%M:%S'`

t1h=`expr $t1 : '\(..\):.*'`
t2h=`expr $t2 : '\(..\):.*'`
t1m=`expr $t1 : '..:\(..\).*'`
t2m=`expr $t2 : '..:\(..\).*'`
t1s=`expr $t1 : '..:..:\(..\)'`
t2s=`expr $t2 : '..:..:\(..\)'`

hdiff=`expr $t2h - $t1h`
mdiff=`expr $t2m - $t1m`
sdiff=`expr $t2s - $t1s`

if [ $tm1 -gt $tm2 ];then
    hdiff=`expr $hdiff - 1`
    mdiff=`expr $tm1 - $tm2`
fi
if [ $ts1 -gt $ts2 ];then
    mdiff=`expr $mdiff - 1`
    sdiff=`expr $sm1 - $sm2`
fi

if [ $hdiff -lt 10 ];then
    hdiff="0$hdiff"
fi
if [ $mdiff -lt 10 ];then
    mdiff="0$mdiff"
fi
if [ $sdiff -lt 10 ];then
    sdiff="0$sdiff"
fi
echo "Elapsed time $hdiff:$mdiff:$sdiff"