how to compile plain CSS with webpack.mix.js - css

Question

I'm using Laravel 8 PHP framework, I need to compile JS and CSS files on localhost using node. When I'm trying to run npm run dev the error output is:
$ npm run dev
> dev
> npm run development
> development
> mix
[webpack-cli] AssertionError [ERR_ASSERTION]: mix.combine() requires a full output file path as the second argument. Got C:\xampp\htdocs\project\public\css
at Function.combine (C:\xampp\htdocs\project\node_modules\laravel-mix\src\Assert.js:54:9)
at Combine.addTask (C:\xampp\htdocs\project\node_modules\laravel-mix\src\components\Combine.js:49:16)
at Combine.register (C:\xampp\htdocs\project\node_modules\laravel-mix\src\components\Combine.js:29:14)
at Object.components.<computed> [as styles] (C:\xampp\htdocs\project\node_modules\laravel-mix\src\components\ComponentRegistrar.js:163:49)
at Object.<anonymous> (C:\xampp\htdocs\project\webpack.mix.js:17:5)
at Module._compile (node:internal/modules/cjs/loader:1101:14)
at Object.Module._extensions..js (node:internal/modules/cjs/loader:1153:10)
at Module.load (node:internal/modules/cjs/loader:981:32)
at Function.Module._load (node:internal/modules/cjs/loader:822:12)
at Module.require (node:internal/modules/cjs/loader:1005:19) {
generatedMessage: false,
code: 'ERR_ASSERTION',
actual: false,
expected: true,
operator: '=='
}
How I can see, the problem is in the mix file, can the wrong syntax be a problem ?
Here is the code
const mix = require('laravel-mix');
mix.js('resources/js/app.js', 'public/js')
.js('resources/assets/js/bootstrap.js', 'public/js')
.js('resources/assets/js/core.min.js', 'public/js')
.js('resources/assets/js/script.js', 'public/js')
.js('resources/assets/js/slider.js', 'public/js')
.js('resources/assets/js/html5shiv.min.js', 'public/js')
.js('resources/assets/js/jquery-3.6.0.min.js', 'public/js')
.js('resources/assets/js/pointer-events.js', 'public/js');
mix.styles([
'resources/assets/css/bootstrap.css',
'resources/assets/css/bootstrap-grid.css',
'resources/assets/css/bootstrap-utilities.css'
],'public/css');
mix.styles([
'resources/assets/css/app.css',
'resources/assets/css/util.css'
], 'public/css');
For this I'm using Laravel 7 Docs because I didn't find anything in Laravel 8 Docs
I'm not using sass or scss, because the source project has ~30000 css lines each file, an I have no idea how I can convert it.
Here is the package.json file
{
"dependencies": {
"bootstrap-icons": "^1.7.2",
"cross-env": "^7.0.3",
"laravel-mix": "^6.0.39"
},
"scripts": {
"dev": "npm run development",
"development": "mix",
"watch": "mix watch",
"watch-poll": "mix watch -- --watch-options-poll=1000",
"hot": "mix watch --hot",
"prod": "npm run production",
"production": "mix --production"
}
}

Answer 1

You will have to pass a file as second argument instead of a folder. So:
mix.styles([
'resources/assets/css/bootstrap.css',
'resources/assets/css/bootstrap-grid.css',
'resources/assets/css/bootstrap-utilities.css'
],'public/css/bootstrap.css');
mix.styles([
'resources/assets/css/app.css',
'resources/assets/css/util.css'
], 'public/css/app.css');