Download File | Hbuz44wwr60l.mp4

import android.content.Intent import android.net.Uri

if __name__ == '__main__': app.run(debug=True) In a Node.js environment with Express, you could achieve this as follows:

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: Download File hbuz44wwr60l.mp4

const express = require('express'); const app = express(); const fs = require('fs'); const path = require('path');

const filePath = '/path/to/your/files/'; // Ensure this is a server-safe path import android

app = Flask(__name__) file_path = '/path/to/your/files/' # Ensure this is a server-safe path

from flask import Flask, send_from_directory console.log('Server listening on port 3000'))

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()