I am a new to gulp and I am working in Gulpfile.js where I have this so far
var gulp = require('gulp');
var sass = require('gulp-sass');
var plumber = require('gulp-plumber');
gulp.task('sass', function() {
gulp.src('public/stylesheets/style.scss')
.pipe(plumber())
.pipe(sass())
.pipe(gulp.dest('public/stylesheets'));
});
gulp.task('watch', function() {
gulp.watch('public/stylesheets/*.scss', ['sass']);
});
gulp.task('default', ['sass', 'watch']);
when I run $ gulp
nothing happens, and I want a gulp file which responds to the gulp command and opens the page in the browser.
There is the I have my folders:
Which is the proper task to accomplish what I want?