elapsed function in bash to print how long things take

12 December 2018   0 comments   Linux, MacOSX

I needed this for a project and it has served me pretty well. Let's jump right into it:

# This is elapsed.sh

SECONDS=0

function elapsed()
{
  local T=$SECONDS
  local D=$((T/60/60/24))
  local H=$((T/60/60%24))
  local M=$((T/60%60))
  local S=$((T%60))
  (( $D > 0 )) && printf '%d days ' $D
  (( $H > 0 )) && printf '%d hours ' $H
  (( $M > 0 )) && printf '%d minutes ' $M
  (( $D > 0 || $H > 0 || $M > 0 )) && printf 'and '
  printf '%d seconds\n' $S
}

And here's how you use it:

# Assume elapsed.sh to be in the current working directory
source elapsed.sh

echo "Doing some stuff..."
# Imagine it does something slow that
# takes about 3 seconds to complete.
sleep 3
elapsed

echo "Some quick stuff..."
sleep 1
elapsed

echo "Doing some slow stuff..."
sleep 61
elapsed

The output of running that is:

Doing some stuff...
3 seconds
Some quick stuff...
4 seconds
Doing some slow stuff...
1 minutes and 5 seconds

Basically, if you have a bash script that does a bunch of slow things, it having a like of elapsed there after some blocks of code will print out how long the script has been running.

It's not beautiful but it works.

Comments

Your email will never ever be published

Related posts

Previous:
How I performance test PostgreSQL locally on macOS 10 December 2018
Next:
Concurrent download with hashin without --update-all 18 December 2018
Related by Keyword:
How to NOT start two servers on the same port 11 June 2018
How I found out where a bash alias was set up 09 May 2018
Run something forever in bash until you want to stop it 13 February 2018
A neat trick to zip a git repo with a version number 01 September 2017
And bash basics 16 October 2015