What is it

notify_after_command_executed is inspired by this Q/A

The purpose of this scripts is to notify about finishing long-running terminal commands.

This script is playing sound and showing notification after execution.

Script is based on https://github.com/rcaloras/bash-preexec

Should work on Debian and Ubuntu

Script depends on

pulseaudio-utils	->	sudo apt-get install pulseaudio-utils
libnotify-bin	->	sudo apt-get install libnotify-bin

Quick test

wget -O - 'https://raw.githubusercontent.com/c0rp-aubakirov/notify_after_command_executed/master/postexec_notify' | bash

Installation

Download

or just clone repo

 $ git clone git@github.com:c0rp-aubakirov/notify_after_command_executed.git

Then add sourcing of script postexec_notify to your .bashrc.

Here is example of installation:

 $ cd ~
 $ git clone git@github.com:c0rp-aubakirov/notify_after_command_executed.git
 $ cd ~/notify_after_command_executed
 $ echo "source $(pwd)/postexec_notify" >> ~/.bashrc

Now restart your terminal. After restart you should see this messages:

preexec.sh is not exist
Trying to download it from github

Now try to test if notification appear:

$ sleep 6

Configuration

There is variable NOTIFICATION_SECONDS_TRESHOLD in postexec_notify script. This variable indicates how long command should work to be considered as long-running. By default it is equals to 5 seconds, this means that any command running more than 5 seconds is considered as long-running command, notification would appear after finish.

I add variable NOTIFICATION_IGNORE_REGEX to ignore notification.(fx-kirin)

Uninstall

Manual

Open your ~/.bashrc file and find there sourcing of postexec_notify. It should look like:

source /home/c0rp/postexec_notify

Remove this line from .bashrc.

Oneline

Here is more automated way:

$ sed -i '/postexec_notify/d' ~/.bashrc