Compiling files with laravel + elixir

Question

I am taking over a laravel project that uses elixir to compile various sass, coffeescript, and javascript files:

elixir(function(mix) {
    mix.sass('main.scss')
        .coffee(['methods.coffee', 'details.coffee', 'cars.coffee', 'context-menu.coffee', 'content.coffee', 'projects.coffee', 'main.coffee'])
        .styles(['main.css'], 'public/css/all.css', 'public/css')
        .scripts(['app.js'], 'public/js/all.js', 'public/js')
        .version(['css/all.css', 'js/all.js']);
});

My questions:

What is the destination of the files below after they are compiled?

.coffee(['methods.coffee', 'details.coffee', 'cars.coffee', 'context-menu.coffee', 'content.coffee', 'projects.coffee', 'main.coffee'])

Does the line below compile the app.js file as all.js and store it in public/js?

.scripts(['app.js'], 'public/js/all.js', 'public/js')

Similarly, does the line below compile the main.css file as all.css and store it in public/css?

.styles(['main.css'], 'public/css/all.css', 'public/css')

Thanks in advance!

Alfredo EM · Accepted Answer · 2016-07-13 16:00:56Z

2

By default, the task will place the compiled in:

sass = public/css/app.css
coffee = public/js/app.js
scripts= public/js/all.js
styles = public/css/all.css

UPDATE

To change the output path just pass a second argument to the method

e.g.

mix.styles([
    'bundle.css'
], 'public/css/aloha');

edited Jul 13, 2016 at 16:00

answered Jul 13, 2016 at 15:52

Alfredo EM

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2 Comments

Trung Tran Over a year ago

Thanks - so to confirm, .styles(['main.css'], 'public/css/all.css', 'public/css') places changes main.css to all.css and places it in public/css correct?

Alfredo EM Over a year ago

The second argument is the base output path (public/css/all.css) and the third argument is the base source path (public/css)

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