Bash: calculate the time elapsed between two timestamps

I have written a script that notifies me when a value is not within a given range. All values “out of range” are logged in a set of per day files.

Every line is timestamped in a proprietary reverse way:

Now, I would like to refine the script, and receive notifications just when at least 60 minutes are passed since the last notification for the given out of range value.

I already solved the issue to print the logs in reverse ordered way with:

for i in $(ls -t /var/log/logfolder/*); do zcat $i|tac|grep !!!|grep --color KEYFORVALUE; done

that results in:

20170817041001 - WARNING: KEYFORVALUE=252.36 is not between 225 and 245 (!!!)
20170817040001 - WARNING: KEYFORVALUE=254.35 is not between 225 and 245 (!!!)
20170817035001 - WARNING: KEYFORVALUE=254.55 is not between 225 and 245 (!!!)
20170817034001 - WARNING: KEYFORVALUE=254.58 is not between 225 and 245 (!!!)
20170817033001 - WARNING: KEYFORVALUE=255.32 is not between 225 and 245 (!!!)
20170817032001 - WARNING: KEYFORVALUE=254.99 is not between 225 and 245 (!!!)
20170817031001 - WARNING: KEYFORVALUE=255.95 is not between 225 and 245 (!!!)
20170817030001 - WARNING: KEYFORVALUE=255.43 is not between 225 and 245 (!!!)
20170817025001 - WARNING: KEYFORVALUE=255.26 is not between 225 and 245 (!!!)
20170817024001 - WARNING: KEYFORVALUE=255.42 is not between 225 and 245 (!!!)
20170817012001 - WARNING: KEYFORVALUE=252.04 is not between 225 and 245 (!!!)

Anyway, I’m stuck at calculating the number of seconds between two of those timestamps, for instance:


What should I do in order to calculate the time elapsed between two timestamps?

Asked By: Marco


With the GNU implementation of date or compatible, this will give you the date in seconds (since the UNIX epoch)

date --date '2017-08-17 04:00:01' +%s    # "1502938801"

And this will give you the date as a readable string from a number of seconds

date --date '@1502938801'    # "17 Aug 2017 04:00:01"

So all that’s needed is to convert your date/timestamp into a format that GNU date can understand, use maths to determine the difference, and output the result


# ksh93 string manipulation (also available in bash, zsh and
# recent versions of mksh)
datestamp1="${datetime1:0:4}-${datetime1:4:2}-${datetime1:6:2} ${datetime1:8:2}:${datetime1:10:2}:${datetime1:12:2}"
datestamp2="${datetime2:0:4}-${datetime2:4:2}-${datetime2:6:2} ${datetime2:8:2}:${datetime2:10:2}:${datetime2:12:2}"

# otherwise use sed
# datestamp1=$(echo "$datetime1" | sed -nE 's/(....)(..)(..)(..)(..)(..)/1-2-3 4:5:6/p')
# datestamp2=$(echo "$datetime2" | sed -nE 's/(....)(..)(..)(..)(..)(..)/1-2-3 4:5:6/p')

seconds1=$(date --date "$datestamp1" +%s)
seconds2=$(date --date "$datestamp2" +%s)

# standard sh integer arithmetics
delta=$((seconds1 - seconds2))
echo "$delta seconds"    # "45197940 seconds"

We’ve not provided timezone information here so it assumes local timezone. Your values for the seconds from the datetime will probably be different to mine. (If your values are UTC then you can use date --utc.)

Answered By: roaima

This is easy with datediff command provided in dateutils package.

datediff -i '%Y%m%d%H%M%S' 20170817040001 20160312000101

See the download page fotlr the latest package and installation file to how to install.

See also project homepage for other distributions installation.

Answered By: αғsнιη

Use the expr command, like:

expr 20170817040001 - 20160312000101

Then you have difference between two values in seconds:

expr 20170817040001 - 20160312000101
Answered By: Jakub