Uglifyjs node js - node.js

Question

This is the code written by me to get all the js files in a directory to be minified:
var http = require('http');
var testFolder = './tests/';
var UglifyJS = require("uglify-js");
var fs = require('fs');
var glob = require("glob");
var fillnam="";
hello();
function hello()
{
glob("gen/*.js", function (er, files) {
//console.log(files);
for(var i=0;i<files.length;i++)
{
fillnam=files[i];
console.log("File Name "+fillnam);
fs.readFile(fillnam, 'utf8', function (err,data)
{
if (err) {
console.log(err);
}
console.log(fillnam+" "+data);
var result = UglifyJS.minify(data);
var gtemp_file=fillnam.replace(".js","");
console.log(gtemp_file);
fs.writeFile(gtemp_file+".min.js", result.code, function(err) {
if(err) {
console.log(err);
} else {
console.log("File was successfully saved.");
}
});
});
}
});
}
http.createServer(function (req, res) {
res.writeHead(200, {'Content-Type': 'text/html'});
res.end('Hello World!');
}).listen(8080);
As a result respective minified js files with same name with .min.js should be formed in the same directory.
But what I am getting is a single file with all files data over written. Like for example if there are two files in a directory a.js and b.js with content:
var a=10;var b=20;
var name="stack";
What I'm getting is single file a.min.js with file content:
var a=10tack;
Please help.

Answer 1

You need to collect all file contents first, concat them and then run UglifyJS.minify on them to be able to save it as a single file.
Something like this (not fully tested)
const testFolder = './tests/';
const UglifyJS = require("uglify-js");
const fs = require('fs');
const readFile = require('util').promisify(fs.readFile);
const glob = require("glob");
function hello() {
glob("gen/*.js", async(er, files) {
let data = [];
for (const file of files) {
const fileData = await readFile(file, {
encoding: 'utf-8'
});
data.push(fileData);
}
const uglified = UglifyJS.minify(data.join('\n'));
fs.writeFile('main.min.js', uglified);
});
}
hello();