How to make a task in gulp automatically run when the project file changes?

Question

JavaScript

I want that when saving changes in the less file, css is automatically collected from it, so as not to write the launch of this action in the console. How to do?
I'm not a programmer, I can't read the documentation, I learned German, don't kick me hard.

Answer 1

It will suit you

gulp.task('watch', function () {
  gulp.watch('assets/less/**/*.less', ['go']);
});

Answer 2

I studied from this article there about what you need and some more useful TYTS
Plus there are automatic compilers for code editors. Those. css is compiled in the folder with the Less file. For example for sublime text - Less2Css