Gulp.js task, return on src?

Question

I\'m new to gulp and have been looking through example set-ups. Some people have the following structure:

gulp.task(\"XXXX\", function() {
    gulp.src(\"...         


        

Answer 1

I found this helpful, if you have multiple streams per task. You need to combine/merge the multiple streams and return them.

var gulp = require('gulp');
var merge = require('gulp-merge');

gulp.task('test', function() {
    var bootstrap = gulp.src('bootstrap/js/*.js')
        .pipe(gulp.dest('public/bootstrap'));

    var jquery = gulp.src('jquery.cookie/jquery.cookie.js')
        .pipe(gulp.dest('public/jquery'));

    return merge(bootstrap, jquery);
});

The alternative, using Gulps task definition structure, would be:

var gulp = require('gulp');

gulp.task('bootstrap', function() {
    return gulp.src('bootstrap/js/*.js')
        .pipe(gulp.dest('public/bootstrap'));
});

gulp.task('jquery', function() {
    return gulp.src('jquery.cookie/jquery.cookie.js')
        .pipe(gulp.dest('public/jquery'));
});

gulp.task('test', ['bootstrap', 'jquery']);