Using requirejs optimizer node module with Gulp

Question

There\'s gulp-requirejs plugin, but it\'s blacklisted with the following message: "use the require.js module directly".

The docs are quite sparse, how would I b

Answer 1

To do this without invoking the shell, you can do it like:

var gulp = require('gulp'),
    rjs = require('requirejs'),

    config = {
        baseUrl: '../appDir/scripts',
        name: 'main',
        out: '../build/main-built.js'
    };

gulp.task('scripts', function(cb){
    rjs.optimize(config, function(buildResponse){
        // console.log('build response', buildResponse);
        cb();
    }, cb);
});

see: https://github.com/phated/requirejs-example-gulpfile/blob/master/gulpfile.js

Answer 2

I just ended up running the r.js build command with gulp-shell:

var gulp = require('gulp'),
    shell = require('gulp-shell');

// Run the r.js command, so simple taks :)
gulp.task('scripts', shell.task([
    'r.js -o build/r/build.js'
]))

It takes roughly 2 seconds to run, not too long at all.

The r settings are defined in an external build.js file that Gulp knows nothing about.