gulp

I have something like this: gulp.task('default', ['css', 'browser-sync'] , function() { gulp.watch(['sass/**/*.scss', 'layouts/*.css'], function() { gulp.run('css'); }); }); but it does not work, because it watches two directories, the sass and the layouts directory for changes. How do I make it work, so that gulp watches anything that happens inside those directories? gulp.task('default', ['css', 'browser-sync'] , function() { gulp.watch(['sass/**/*.scss', 'layouts/**/*.css'], ['css']); }); sass/**/*.scss and layouts/**/*.css will watch every directory and subdirectory for any changes to

I am new to Gulp and wondering how can I minify (concatinate) my collection to single file instead of every file separately. Example var gulp = require('gulp'); var uglify = require('gulp-uglify'); gulp.task('minify', function () { gulp.src([ 'content/plugins/jquery/jquery-2.1.1.js', 'content/plugins/jquery/jquery-ui.js', 'content/js/client.js']) .pipe(uglify()) .pipe(gulp.dest('content/js/client.min.js')) // It will create folder client.min.js }); gulp.task('default', ['minify']); As you can see I'm trying to minify 3 javascript files and then output them into one client.min.js . Is it

I tried to install gulp by npm install -g gulp The output seems to be something like this. (i have skipped some logs) npm http 304 https://registry.npmjs.org/string_decoder npm http 304 https://registry.npmjs.org/lodash._htmlescapes /usr/bin/gulp -> /usr/lib/node_modules/gulp/bin/gulp.js gulp@3.6.2 /usr/lib/node_modules/gulp ├── tildify@0.2.0 ├── pretty-hrtime@0.2.1 ├── deprecated@0.0.1 ├── archy@0.0.2 ├── semver@2.2.1 ├── chalk@0.4.0 (has-color@0.1.7, ansi-styles@1.0.0, strip-ansi@0.1.1) ├── orchestrator@0.3.3 (sequencify@0.0.7) ├── liftoff@0.9.8 (extend@1.2.1, minimist@0.0.8, resolve@0.6.3,

I am trying to use gulp-requirejs to build a demo project. I expect result to be a single file with all js dependencies and template included. Here is my gulpfile.js var gulp = require('gulp'); var rjs = require('gulp-requirejs'); var paths = { scripts: ['app/**/*.js'], images: 'app/img/**/*' }; gulp.task('requirejsBuild', function() { rjs({ name: 'main', baseUrl: './app', out: 'result.js' }) .pipe(gulp.dest('app/dist')); }); // The default task (called when you run `gulp` from cli) gulp.task('default', ['requirejsBuild']); The above build file works with no error, but the result.js only

I have a project that has: JS client with somewhat interesting build process. That includes compiling CSS, catenating and minifying JS and CSS, generating and processing HTML, and some other steps. The Node tools like Grunt or Gulp are great at this. Java server that is a WAR deployed on Tomcat. It includes these assets as well as all the Java code. It has all kinds of tests: Unit tests, integration tests that may instantiate a DAO and talk to DB, and end-to-end API tests that actually talk to the app running on Tomcat. End-to-end tests with Protractor. If you're not familiar, it's another