gulp loop over json object

Question

I have all of my paths defined at the top of the project,

var paths = {
  scripts: ['./js/main.js', './js/datetimepicker.js'],
  styles: ['./stylesheets/style.scss', './stylesheets/datetimepicker/datetimepicker.css', './stylesheets/vis/vis.css'],
  images: ['img/**/*'],
  libs: ['./js/*.min.js', './img/*', './fonts/FontAwesome/*'],
  compress: ['./css/**', './fonts/FontAwesome/**', './img/**', './js/main.min.js', './js/bootstrap.min.js'],
  concat: [{
    'src': './web-src/js/services*.js',
    'name': 'services.js'
  }]
};

My issue is that when I call paths.concat[0].src I am forced to specify what index in the array I want. So when I call my concat task,

gulp.task('concat', function () {
  return gulp.src(paths.concat[0].src)
    .pipe(concat(paths.concat[0].src))
    .pipe(gulp.dest('./dist/'));
});

Having the paths defined is not helpful. What is the 'best practice' way of looping over a task?

Thanks

Probably not the correct name but I'm assigning the json object to a variable. — Grady D
– Grady D, Commented Dec 31, 2015 at 3:30

Grady D · Accepted Answer · 2015-12-31 21:11:08Z

2

I accomplished this using,

gulp.task('concat', function () {
    return paths.concat.forEach(function(obj) {
        return gulp.src(obj.src)
            .pipe(concat(obj.name))
            .pipe(rename(function (path) {
                path.basename += '.min';
                return path;
            }))
            .pipe(gulp.dest('./web/js'));
    });
});

answered Dec 31, 2015 at 21:11

Grady D

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