Express: Minify CSS/JS from remote URL with UglifyJS - node.js

Question

I managed to get the data on #1, but I can't print it out in Express #2, how can I resolve the data and output it?
I believe this is my code error where I can't place the correct async/await.
Here's my code:
minify.js
const { getContent } = require('./get');
const { minifyJS } = require('./processing');
async function getMinify(req, res) {
try {
const remoteUrl = 'http://example.com/script.js';
console.log('URL: ' + remoteUrl);
// #1
const content = await getContent(remoteUrl);
// #2
const outputJS = await minifyJS(content);
res.end(outputJS);
} catch (error) {
res.end(error.content);
}
}
module.exports = { getMinify }
get.js
const got = require('got');
async function getContent(remoteUrl) {
const code = await got(remoteUrl);
return code.body;
}
module.exports = { getContent }
processing.js
const { minify } = require('uglify-js');
async function minifyJS(data) {
const result = await minify(data);
return result.code;
}
module.exports = { minifyJS }
app.js
const express = require('express');
const { getMinify } = require('./minify');
const app = express();
app.get('/*', getMinify);
app.listen(5000);

Answer 1

based on uglify-js, result of minify(data) function has two object code and error.
and minify(data) don't need to await
processing.js should changed
const { minify } = require("uglify-js");
async function minifyJS(data) {
let { error, code } = minify(data);
if (error)
throw error;
return code;
}
module.exports = { minifyJS };
and for handling of two .js file (get.js and processing.js) changed
minify.js
try {
const remoteUrl = 'http://example.com/script.js';
console.log('URL: ' + remoteUrl);
// #1
const content = await getContent(remoteUrl);
// #2
const outputJS = await minifyJS(content);
res.end(outputJS);
} catch (error) {
res.end(error); // remove .content
}