Subtracting two timestamps in bash script

If you want to process the date using simple command line tools, you need to convert the timestamps into some easy-to-deal-with format, like epoch-based.

You can do this with the date command, which you can wrap in a function like so:

timestamp() {     date '+%s%N' --date="$1"}# timestamp '2020-01-01 12:20:45.12345' => 1577899245123450000

Then you can subtract them directly:

echo $(( $(timestamp "$etime") - $(timestamp "$stime") ))

Note that the %N format specifier (nanoseconds) is a GNU extension and is not handled in the default macOS date command. Either (a) brew install coreutils and use gdate in the function above or (b) use this alternative function on macOS (but note it lacks support for sub-second measurements):

timestamp() {    local format='%Y-%m-%d %H:%M:%S'     # set to whatever format is used    date -j -f "$format" "$1" '+%s'}# timestamp '2020-01-01 12:20:45' => 1577899245

If you need to use bash, you need to convert the timestamp into an integer value:

$ time="12:34:56.789"$ IFS=":." read -r h m s ms <<<"$time"$ echo $h $m $s $ms12 34 56 789$ milliseconds=$(( (h*3600 + m*60 + s)*1000 + ms ))$ echo $milliseconds45296789

Then you can subtract to get a number of milliseconds representing the diff.

Convert to seconds with some arithmetic and string construction:

$ seconds="$((milliseconds / 1000)).$((milliseconds % 1000))"$ echo $seconds45296.789

To address gniourf_gniourf's valid point (in a clunky way):

$ time="12:34:56.7"$ IFS=":." read -r h m s ms <<<"$time"$ ms="${ms}000"; ms=${ms:0:3}$ echo $h $m $s $ms12 34 56 700# .......^^^

Also, strings that are invalid octal numbers like "08" and "09" will throw errors in bash arithmetic, so explicitly declare they are base 10

milliseconds=$(( (10#$h*3600 + 10#$m*60 + 10#$s)*1000 + 10#$ms ))

Assuming your variables are integers, you need to enclose your calculation in an arithmetic evaluation for that to work.

echo $(( VAR1 - VAR2 ))

Or, if you want to assign the value :

VAR3=$(( VAR1 - VAR2 ))