Compute time-difference on millisecond level - bash

Question

Assuming the following time-formats:
MM:DD:YYYY hh:mm:ss:nn
How can I compute the difference between two times? I have tried the
following, but it seems to fail.
% Value1='08:27:2018 23:53:50:08'
% Value2='08:28:2018 00:00:08:89'
% echo "$(($(date -d "$Value2" '+%s') - $(date -d "$Value1" '+%s')))"

Answer 1

Update: as the OP changed his format
$ Value1='08:27:2018 23:53:50:08'
$ Value2='08:28:2018 00:00:08:89'
$ Value1=${Value1/://}; Value1=${Value1/://}; Value1=${Value1%:*}.${Value1##*:}
$ Value2=${Value2/://}; Value2=${Value2/://}; Value2=${Value2%:*}.${Value2##*:}
$ echo $(date -d "$Value2" '+%s.%N') - $(date -d "$Value1" '+%s.%N') | bc -l
378.810000000
So all you need to do is convert it to a format that date knows, this is MM/DD/YYYY
Original answer below:
The problem is that your date-time format is not really recognized.
The date format:
$ date -d "08272018"
date: invalid date ‘08272018’
The date command knows many formats, but it is hard for it to distinguish between YYYYMMDD, DDMMYYYY and MMDDYYYY. To be clear, what does 10021002 represent as a date?
In this simple format—without delimiters—date will recognize YYYYMMDD and YYMMDD
$ date -d "20180708"
Sun 8 Jul 00:00:00 UTC 2018
The time format:
The notation HH:MM:SS:ss is by far from standard. What does ss represent, ss 60th of a second? The normal notation would be more HH:MM:SS.sss This will be recognized.
$ date -d "23:53:50:08"
date: invalid date ‘23:53:50:08’
$ date -d "23:53:50.08" "+%a %d %b %T.%N %Z %Y"
Wed 12 Sep 23:53:50.080000000 UTC 2018
So if you get your date format correct, you already get a long way:
% Value1='20180827 23:53:50.08'
% Value2='20180828 00:00:08.89'
% echo "$(($(date -d "$Value2" '+%s') - $(date -d "$Value1" '+%s')))"
378
The sad thing is that we are missing our milliseconds for this you need floating point arithmetic and bash does not support it. But there are ways around that (How do I use floating-point division in bash?)
$ echo $(date -d "$Value2" '+%s.%N') - $(date -d "$Value1" '+%s.%N') | bc -l
378.810000000

Answer 2

I shortened variable names:
v1='08:27:2018 23:53:50:08'
v2='08:28:2018 00:00:08:89'
With GNU date, just stick to one safe input format, you can convert YYYY-MM-DD HH:MM:SS.NN to... anything another. (side note: I love freebsd date, where you can just specify -f option for strptime. I wish I could do that with GNU date). So we can:
v1_epoch=$(date -d "${v1:6:4}-${v1:0:2}-${v1:3:2} ${v1:11:2}:${v1:14:2}:${v1:17:2}.${v1:20}" +%s.%N)
v2_epoch=$(date -d "${v2:6:4}-${v2:0:2}-${v2:3:2} ${v2:11:2}:${v2:14:2}:${v2:17:2}.${v2:20}" +%s.%N)
It will get us values of seconds with nanosecond resolution since epoch time. Now we need to calc a difference, we need to use a tool like bc, cause bash does not support floating point calculations.
diff=$(printf "scale=9; %s - %s\n" "$v2_epoch" "$v1_epoch" | bc)
Now this represents the difference of time we need to represent in hours, minutes, seconds and miliseconds.
printf "%s.%.3s" $(date -d#"$diff" -u +'%H:%M:%S %N')
That's simple, but it will wrap around at 23 hours, so we can do better with bc. The rounding in bc is sometimes unexpected... you need to just get used to unexpected scale=0 lines:
printf "%02d:%02d:%02d.%03d\n" $(printf 'scale=11; a=%s; scale=0; h=a/3600; m=a%%3600/60; s=a%%60/1; ms=a*1000%%1000/1; h \n m \n s \n ms \n' '$diff' | bc -l)
A "neat" oneliner:
$ v1='08:27:2018 23:53:50:08'
$ v2='08:28:2018 00:00:08:89'
$ printf "%02d:%02d:%02d.%03d\n" $(printf 'scale=11; a=%s; scale=0; h=a/3600; m=a%%3600/60; s=a%%60/1; ms=a*1000%%1000/1; h \n m \n s \n ms \n' "$(printf "scale=9; %s - %s\n" "$(date -d "${v2:6:4}-${v2:0:2}-${v2:3:2} ${v2:11:2}:${v2:14:2}:${v2:17:2}.${v2:20}" +%s.%N)" "$(date -d "${v1:6:4}-${v1:0:2}-${v1:3:2} ${v1:11:2}:${v1:14:2}:${v1:17:2}.${v1:20}" +%s.%N)" | bc)" | bc -l)
I guess this could be even shortened with some here strings, but that just harms readability:
printf "%02d:%02d:%02d.%03d\n" $(<<<"scale=11; a=$(<<< "scale=9; $(date -d "${v2:6:4}-${v2:0:2}-${v2:3:2} ${v2:11:2}:${v2:14:2}:${v2:17:2}.${v2:20}" +%s.%N) - $(date -d "${v1:6:4}-${v1:0:2}-${v1:3:2} ${v1:11:2}:${v1:14:2}:${v1:17:2}.${v1:20}" +%s.%N)"$'\n' bc); scale=0; h=a/3600; m=a%3600/60; s=a%60/1; ms=a*1000%1000/1; h"$'\n'"m"$'\n'"s"$'\n'"ms"$'\n' bc -l)
Or you can create a function for conversion:
mydate_read() { date -d "${1:6:4}-${1:0:2}-${1:3:2} ${1:11:2}:${1:14:2}:${1:17:2}.${1:20}" +%s.%N; };
printf "%02d:%02d:%02d.%03d\n" $(<<<"scale=11; a=$(<<< "scale=9; $(mydate_read "$v2") - $(mydate_read "$v1")"$'\n' bc); scale=0; h=a/3600; m=a%3600/60; s=a%60/1; ms=a*1000%1000/1; h"$'\n'"m"$'\n'"s"$'\n'"ms"$'\n' bc -l)
I forgot, we can merge the two bc calls into one:
mydate_read() { date -d "${1:6:4}-${1:0:2}-${1:3:2} ${1:11:2}:${1:14:2}:${1:17:2}.${1:20}" +%s.%N; };
printf "%02d:%02d:%02d.%03d\n" $(printf 'scale=11; a=%s - %s; scale=0; h=a/3600; m=a%%3600/60; s=a%%60/1; ms=a*1000%%1000/1; h \n m \n s \n ms \n' "$(mydate_read "$v2")" "$(mydate_read "$v1")" | bc -l)