Concat scripts in order with Gulp

Question

Say, for example, you are building a project on Backbone or whatever and you need to load scripts in a certain order, e.g. underscore.js needs to be loaded befo

Answer 1

The sort-stream may also be used to ensure specific order of files with gulp.src. Sample code that puts the backbone.js always as the last file to process:

var gulp = require('gulp');
var sort = require('sort-stream');
gulp.task('scripts', function() {
gulp.src(['./source/js/*.js', './source/js/**/*.js'])
  .pipe(sort(function(a, b){
    aScore = a.path.match(/backbone.js$/) ? 1 : 0;
    bScore = b.path.match(/backbone.js$/) ? 1 : 0;
    return aScore - bScore;
  }))
  .pipe(concat('script.js'))
  .pipe(stripDebug())
  .pipe(uglify())
  .pipe(gulp.dest('./build/js/'));
});