I am a new to using Gulp, just trying to learn it...Now the problem i get and want to ask is the way to setup default task with watch and browser sync included I need to know am i doing something wrong
Can anybody improve my code here, i don't understand the relation of watch and browser sync, which tasks to run before browser-sync and when to watch
Below is my folder structure
var gulp = require('gulp');
var browserSync = require('browser-sync');
var reload = browserSync.reload;
var uglify = require('gulp-uglify');
var less = require('gulp-less');
var plumber= require('gulp-plumber');
var cssmin = require('gulp-cssmin');
var rename = require('gulp-rename');
var htmlmin = require('gulp-htmlmin');
var imagemin = require ('gulp-imagemin');
//scripts task
//uglifies
gulp.task('scripts', function(){
gulp.src('js/*.js')
.pipe(plumber())
.pipe(uglify())
.pipe(gulp.dest('build/js'));
});
//compress images
gulp.task('imagemin', function(){
gulp.src('img/**/*.+(png|jpg|gif|svg)')
.pipe(cache(imagemin({
interlaced: true
})))
.pipe(gulp.dest('build/img'));
});
//CSS styles
gulp.task('less', function(){
gulp.src('less/style.less')
.pipe(plumber())
.pipe(less())
.pipe(gulp.dest('build/css'));
});
gulp.task('cssmin', function(){
gulp.src('build/css/style.css')
.pipe(plumber())
.pipe(cssmin())
.pipe(rename({suffix: '.min'}))
.pipe(gulp.dest('build/css'))
.pipe(reload({stream:true})); // inject into browsers
});
gulp.task('htmlmin', function(){
return gulp.src('*.html')
.pipe(htmlmin({removeComments: true}))
.pipe(gulp.dest('build'))
.pipe(reload({stream:true})); // inject into browsers
});
// Browser-sync task, only cares about compiled CSS
gulp.task('browser-sync', function() {
browserSync(['css/*.css', 'js/*.js','less/*.less', 'images/*'],{
server: {
baseDir: "./"
}
});
});
/* Watch scss, js and html files, doing different things with each. */
gulp.task('default', ['browser-sync' , 'scripts', 'less', 'cssmin', 'htmlmin', 'imagemin'], function () {
/* Watch scss, run the sass task on change. */
gulp.watch(['less/**/*.less'], ['less'])
//Watch css min
gulp.watch(['build/css/*.css'], ['cssmin'])
/* Watch app.js file, run the scripts task on change. */
gulp.watch(['js/*.js'], ['scripts'])
/* Watch .html files, run the bs-reload task on change. */
gulp.watch(['*.html'], ['htmlmin']);
// gulp.watch('app/*.html', browser-sync.reload);
// gulp.watch('app/js/**/*.js', browser-sync.reload);
});
Now the process i want is
I would not worry about minification at this point if your goal is to run in development mode, this applies for imagemin (i would do that offline anyways), cssmin, htmlmin, and your js task that runs uglify by default. Ideally you would want to debug in the browser, and having your code minified will not help you much. If you add a dist
task to perform the minification step.
I understand that you need Less to CSS for obvious reasons. So you are looking for something like this:
var gulp = require('gulp');
var plumber = require('gulp-plumber');
var browserSync = require('browser-sync').create();
var sass = require('gulp-less');
// Static Server + watching less/html files
gulp.task('serve', ['less'], function() {
browserSync.init({
server: "./"
});
gulp.watch("less/*.less", ['less']);
gulp.watch("*.html").on('change', browserSync.reload);
});
gulp.task('less', function(){
gulp.src('less/style.less')
.pipe(plumber())
.pipe(less())
.pipe(gulp.dest('build/css'))
.pipe(browserSync.stream());
});
gulp.task('default', ['serve']);
This code invokes serve
as the main task. Serve
task has less
as a dependency (which is going to be invoked first). Then, the callback is finally invoked. BrowserSync is initialized and a watch is added for both html files and less files.
Check out this page if you want to learn more about gulp + browsersync integration.