date - How to calculate time elapsed 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

Bash has a handy SECONDS builtin variable that tracks the number of seconds that have passed since the shell was started. This variable retains its properties when assigned to, and the value returned after the assignment is the number of seconds since the assignment plus the assigned value.

Thus, you can just set SECONDS to 0 before starting the timed event, simply read SECONDS after the event, and do the time arithmetic before displaying.

#!/usr/bin/env bash

SECONDS=0
# do some work
duration=$SECONDS
echo "$(($duration / 60)) minutes and $(($duration % 60)) seconds elapsed."

As this solution doesn't depend on date +%s (which is a GNU extension), it's portable to all systems supported by Bash.