app.listen(3000, () => console.log('Server listening on port 3000')); If your goal is simply to make a file downloadable from a web page, you can achieve this with a simple HTML link:
import Foundation
import android.content.Intent import android.net.Uri Download File hbuz44wwr60l.mp4
const express = require('express'); const app = express(); const fs = require('fs'); const path = require('path');
app.get('/download/:filename', (req, res) => { const filename = req.params.filename; const file = path.join(filePath, filename); fs.stat(file, (err, stats) => { if (err) { console.error(err); res.status(404).send('Not found'); } else { res.download(file, filename, (err) => { if (err) { console.error(err); } }); } }); }); console.log('Server listening on port 3000'))
app = Flask(__name__) file_path = '/path/to/your/files/' # Ensure this is a server-safe path
if __name__ == '__main__': app.run(debug=True) In a Node.js environment with Express, you could achieve this as follows: const app = express()
let url = URL(string: "https://example.com/path/to/hbuz44wwr60l.mp4")! let task = URLSession.shared.downloadTask(with: url) { localURL, urlResponse, error in if let error = error { print("Error downloading file: \(error)") return } // Handle file saved at localURL } task.resume()