How to calculate time difference in bash script?

Question

I print the start and end time using date +"%T", which results in something like:

10:33:56
10:36:10

How could I calculate and print the difference between these two?

I would like to get something like:

2m 14s

Answer 1

Another option is to use datediff from dateutils (http://www.fresse.org/dateutils/#datediff):

$ datediff 10:33:56 10:36:10
134s
$ datediff 10:33:56 10:36:10 -f%H:%M:%S
0:2:14
$ datediff 10:33:56 10:36:10 -f%0H:%0M:%0S
00:02:14

You could also use gawk. mawk 1.3.4 also has strftime and mktime but older versions of mawk and nawk don't.

$ TZ=UTC0 awk 'BEGIN{print strftime("%T",mktime("1970 1 1 10 36 10")-mktime("1970 1 1 10 33 56"))}'
00:02:14

Or here's another way to do it with GNU date:

$ date -ud@$(($(date -ud'1970-01-01 10:36:10' +%s)-$(date -ud'1970-01-01 10:33:56' +%s))) +%T
00:02:14

Answer 2

% start=$(date +%s)
% echo "Diff: $(date -d @$(($(date +%s)-$start)) +"%M minutes %S seconds")"
Diff: 00 minutes 11 seconds

Answer 3

Seconds

To measure elapsed time (in seconds) we need:

• an integer that represents the count of elapsed seconds and
• a way to convert such integer to an usable format.

---

An integer value of elapsed seconds:

• There are two bash internal ways to find an integer value for the number of elapsed seconds:

  1. Bash variable SECONDS (if SECONDS is unset it loses its special property).

     • Setting the value of SECONDS to 0:

       SECONDS=0
       sleep 1  # Process to execute
       elapsedseconds=$SECONDS
       

     • Storing the value of the variable SECONDS at the start:

       a=$SECONDS
       sleep 1  # Process to execute
       elapsedseconds=$(( SECONDS - a ))
       

  2. Bash printf option %(datefmt)T:

     a="$(TZ=UTC0 printf '%(%s)T\n' '-1')"    ### `-1`  is the current time
     sleep 1                                  ### Process to execute
     elapsedseconds=$(( $(TZ=UTC0 printf '%(%s)T\n' '-1') - a ))
     

---

Convert such integer to an usable format

The bash internal printf can do that directly:

$ TZ=UTC0 printf '%(%H:%M:%S)T\n' 12345
03:25:45

similarly

$ elapsedseconds=$((12*60+34))
$ TZ=UTC0 printf '%(%H:%M:%S)T\n' "$elapsedseconds"
00:12:34

but this will fail for durations of more than 24 hours, as we actually print a wallclock time, not really a duration:

$ hours=30;mins=12;secs=24
$ elapsedseconds=$(( ((($hours*60)+$mins)*60)+$secs ))
$ TZ=UTC0 printf '%(%H:%M:%S)T\n' "$elapsedseconds"
06:12:24

For the lovers of detail, from bash-hackers.org:

%(FORMAT)T outputs the date-time string resulting from using FORMAT as a format string for strftime(3). The associated argument is the number of seconds since Epoch, or -1 (current time) or -2 (shell startup time). If no corresponding argument is supplied, the current time is used as default.

So you may want to just call textifyDuration $elpasedseconds where textifyDuration is yet another implementation of duration printing:

textifyDuration() {
   local duration=$1
   local shiff=$duration
   local secs=$((shiff % 60));  shiff=$((shiff / 60));
   local mins=$((shiff % 60));  shiff=$((shiff / 60));
   local hours=$shiff
   local splur; if [ $secs  -eq 1 ]; then splur=''; else splur='s'; fi
   local mplur; if [ $mins  -eq 1 ]; then mplur=''; else mplur='s'; fi
   local hplur; if [ $hours -eq 1 ]; then hplur=''; else hplur='s'; fi
   if [[ $hours -gt 0 ]]; then
      txt="$hours hour$hplur, $mins minute$mplur, $secs second$splur"
   elif [[ $mins -gt 0 ]]; then
      txt="$mins minute$mplur, $secs second$splur"
   else
      txt="$secs second$splur"
   fi
   echo "$txt (from $duration seconds)"
}

---

GNU date.

To get formated time we should use an external tool (GNU date) in several ways to get up to almost a year length and including Nanoseconds.

Math inside date.

There is no need for external arithmetic, do it all in one step inside date:

date -u -d "0 $FinalDate seconds - $StartDate seconds" +"%H:%M:%S"

Yes, there is a 0 zero in the command string. It is needed.

That's assuming you could change the date +"%T" command to a date +"%s" command so the values will be stored (printed) in seconds.

Note that the command is limited to:

• Positive values of $StartDate and $FinalDate seconds.
• The value in $FinalDate is bigger (later in time) than $StartDate.
• Time difference smaller than 24 hours.
• You accept an output format with Hours, Minutes and Seconds. Very easy to change.
• It is acceptable to use -u UTC times. To avoid "DST" and local time corrections.

---

If you must use the 10:33:56 string, well, just convert it to seconds,
also, the word seconds could be abbreviated as sec:

string1="10:33:56"
string2="10:36:10"
StartDate=$(date -u -d "$string1" +"%s")
FinalDate=$(date -u -d "$string2" +"%s")
date -u -d "0 $FinalDate sec - $StartDate sec" +"%H:%M:%S"

Note that the seconds time conversion (as presented above) is relative to the start of "this" day (Today).

---

The concept could be extended to nanoseconds, like this:

string1="10:33:56.5400022"
string2="10:36:10.8800056"
StartDate=$(date -u -d "$string1" +"%s.%N")
FinalDate=$(date -u -d "$string2" +"%s.%N")
date -u -d "0 $FinalDate sec - $StartDate sec" +"%H:%M:%S.%N"

---

If is required to calculate longer (up to 364 days) time differences, we must use the start of (some) year as reference and the format value %j (the day number in the year):

Similar to:

string1="+10 days 10:33:56.5400022"
string2="+35 days 10:36:10.8800056"
StartDate=$(date -u -d "2000/1/1 $string1" +"%s.%N")
FinalDate=$(date -u -d "2000/1/1 $string2" +"%s.%N")
date -u -d "2000/1/1 $FinalDate sec - $StartDate sec" +"%j days %H:%M:%S.%N"

Output:
026 days 00:02:14.340003400

Sadly, in this case, we need to manually subtract 1 ONE from the number of days. The date command view the first day of the year as 1. Not that difficult ...

a=( $(date -u -d "2000/1/1 $FinalDate sec - $StartDate sec" +"%j days %H:%M:%S.%N") )
a[0]=$((10#${a[0]}-1)); echo "${a[@]}"

The use of long number of seconds is valid and documented here:
https://www.gnu.org/software/coreutils/manual/html_node/Examples-of-date.html#Examples-of-date

---

Busybox date

A tool used in smaller devices (a very small executable to install): Busybox.

Either make a link to busybox called date:

$ ln -s /bin/busybox date

Use it then by calling this date (place it in a PATH included directory).

Or make an alias like:

$ alias date='busybox date'

---

Busybox date has a nice option: -D to receive the format of the input time. That opens up a lot of formats to be used as time. Using the -D option we can convert the time 10:33:56 directly:

date -D "%H:%M:%S" -d "10:33:56" +"%Y.%m.%d-%H:%M:%S"

And as you can see from the output of the Command above, the day is assumed to be "today". To get the time starting on epoch:

$ string1="10:33:56"
$ date -u -D "%Y.%m.%d-%H:%M:%S" -d "1970.01.01-$string1" +"%Y.%m.%d-%H:%M:%S"
1970.01.01-10:33:56

Busybox date can even receive the time (in the format above) without -D:

$ date -u -d "1970.01.01-$string1" +"%Y.%m.%d-%H:%M:%S"
1970.01.01-10:33:56

And the output format could even be seconds since epoch.

$ date -u -d "1970.01.01-$string1" +"%s"
52436

For both times, and a little bash math (busybox can not do the math, yet):

string1="10:33:56"
string2="10:36:10"
t1=$(date -u -d "1970.01.01-$string1" +"%s")
t2=$(date -u -d "1970.01.01-$string2" +"%s")
echo $(( t2 - t1 ))

Or formatted:

$ date -u -D "%s" -d "$(( t2 - t1 ))" +"%H:%M:%S"
00:02:14

Answer 4

date can give you the difference and format it for you (OS X options shown)

date -ujf%s $(($(date -jf%T "10:36:10" +%s) - $(date -jf%T "10:33:56" +%s))) +%T
# 00:02:14

date -ujf%s $(($(date -jf%T "10:36:10" +%s) - $(date -jf%T "10:33:56" +%s))) \
    +'%-Hh %-Mm %-Ss'
# 0h 2m 14s

Some string processing can remove those empty values

date -ujf%s $(($(date -jf%T "10:36:10" +%s) - $(date -jf%T "10:33:56" +%s))) \
    +'%-Hh %-Mm %-Ss' | sed "s/[[:<:]]0[hms] *//g"
# 2m 14s

This won't work if you place the earlier time first. If you need to handle that, change $(($(date ...) - $(date ...))) to $(echo $(date ...) - $(date ...) | bc | tr -d -)

Answer 5

Here is a solution using only the date commands capabilities using "ago", and not using a second variable to store the finish time:

#!/bin/bash

# save the current time
start_time=$( date +%s.%N )

# tested program
sleep 1

# the current time after the program has finished
# minus the time when we started, in seconds.nanoseconds
elapsed_time=$( date +%s.%N --date="$start_time seconds ago" )

echo elapsed_time: $elapsed_time

this gives:

$ ./time_elapsed.sh 
elapsed_time: 1.002257120

Answer 6

Or wrap it up a bit

alias timerstart='starttime=$(date +"%s")'
alias timerstop='echo seconds=$(($(date +"%s")-$starttime))'

Then this works.

timerstart; sleep 2; timerstop
seconds=2