Laravel Elixir - Execute cmd on watch update - laravel-elixir

Question

I'm trying to create an easy workflow for package development and was hoping someone could pint me in the right direction. In short, once a css file has been updated I want to be able to run a command (php artisan vendor:publish --force) to automatically publish the files.
Can this be done with Elixir and if so could anyone point me in the right direction?
regards

Answer 1

This was incredibly useful to get me on the right track, but I found a slightly simpler solution that I thought I would share:
var elixir = require('laravel-elixir');
var gulp = require("gulp");
var shell = require("gulp-shell");
elixir(function(mix) {
mix.task('publish_assets', ['resources/assets/**/*.scss', 'resources/assets/**/*.js']);
});
gulp.task('publish_assets', shell.task([
"php ../../../../artisan vendor:publish --tag=public --force"
]));
It watches all of the js and sass files for changes and runs the publish assets task. Mine is set to only publish the "public" tagged files, and a quick note as well, you will need to install the gulp-shell utility (https://www.npmjs.com/package/gulp-shell) for either of these.

Answer 2

if anyone is looking, this works nicely:
var gulp = require("gulp");
var shell = require("gulp-shell");
var elixir = require("laravel-elixir");
elixir.extend("publish", function() {
var baseDir = this.assetsDir;
gulp.task("publish_assets", function() {
gulp.src("").pipe( shell( [
"php ../../../artisan vendor:publish --force",
"cd ../../../ ; gulp"
]));
});
this.registerWatcher("publish_assets", [
baseDir + '**/*.scss',
baseDir + '**/*.js'
]);
return this.queueTask("publish_assets");
});