How to make dynamic require (webpack)?

Question

JavaScript

Hello.
Unfortunately, the nerves are already at the limit ..... Please, tell me, please ....
There are two entry points:

entry: {
        Api : './modules/Api/resources/assets/frontend/app/index',
        Test : './modules/Test/resources/assets/frontend/app/index'
    }, 
output: {
        path: path.resolve(__dirname, 'public/js'),
        filename: '[name].js',
        chunkFilename: '[name].js',
        publicPath: '/js/'
    },
resolve : {
        alias : {
            Api : path.resolve(__dirname, 'modules/Api/resources/assets/frontend/app/index'),
            Test : path.resolve(__dirname, 'modules/Test/resources/assets/frontend/app/index')
        }
    }

PS the main thing is to pay attention to the paths in the entry!
From the Api.js module, I want to include Test.js,
BUT with

var name = 'Test';

require.ensure([], require => {
    require("./" + name);
});

Could you please clarify how this can be done?
PS I know there is a context, but alas it doesn't work like this:

var req = require.context('bundle!./../../../../../', false, /^\.\/.*\.js$/);
req("./" + name + "/resources/assets/frontend/app/require");

I tried url:
require("url?./" + name);
it doesn't work...
Alas, yes, it works like this (code from Api.js):

require.ensure([], require => {
        require('./../../../../../Test/resources/assets/frontend/app');
    });

But, and if I delete the folder ("./modules/Test"), then there will be an error during compilation, and I would like to avoid this. No matter how there is no module, well, to hell with it, let what is work.

Answer 1

If you have registered alias, then just write

require.ensure(['Test'], require => {
    require('Test');
});