I want to use the date command to adjust a clock time by adding and subtracting hours, minutes, and seconds, without having time zones affect the calculation. For example, starting with 21:00:00, I want to add 2:02:30 and subtract 1:10, producing 23:01:20. What is the best way to do this?
3 Answers
For portable and predictable arithmetic, convert each value to seconds, perform integer arithmetic, then format the result back into hours, minutes, and seconds. For example: `s=$((3600*(21+2)+60*(0+2-1)+(0+30-10))); printf '%02d:%02d:%02dn' $((s/3600%24)) $((s/60%60)) $((s%60))` prints `23:01:20`. This avoids date parsing and time-zone or daylight-saving surprises.
Another option is to convert the times to Unix timestamps, do the arithmetic, and convert the result back with `date -d @SECONDS +%T`. Set `TZ=UTC` or another fixed-offset zone if you want the calculation to ignore local daylight-saving rules. For example: `TZ=UTC date -d '21:00:00 2 hours 2 minutes 30 seconds 1 minute ago 10 seconds ago' +%T`.
With GNU date, write the adjustments as explicit relative units rather than HH:MM:SS values. This produces the expected result: `date -d '21:00:00 2 hours 2 minutes 30 seconds 1 minute ago 10 seconds ago' +%T` → `23:01:20`. If you mean one minute and ten seconds, write both units explicitly; `-01:10` may be interpreted as hours and minutes instead.
This syntax is GNU date-specific. POSIX date does not provide the same relative-date parsing, so check which implementation is installed before relying on it.