How to use nodejs to open default browser and navigate to a specific URL

Question

I'm writing an application using Node.js.

One of the functions I want to create is to open the default web browser and navigate to a specific URL.

I want it to be portable so that it runs on Windows/Mac/Linux.

Answer 1

Use open (formerly known as opn) because it will handle the cross platform issue. To install:

$ npm install open

To use:

const open = require('open');

// opens the url in the default browser 
open('http://sindresorhus.com');
 
// specify the app to open in 
open('http://sindresorhus.com', {app: 'firefox'});

Answer 2

var url = 'http://localhost';
var start = (process.platform == 'darwin'? 'open': process.platform == 'win32'? 'start': 'xdg-open');
require('child_process').exec(start + ' ' + url);

Answer 3

node-open is deprecated. Now use open:

const open = require('open')

await open('http://sindresorhus.com') // Opens the url in the default browser

await open('http://sindresorhus.com', {app: 'firefox'}) // Specify the app to open in

Answer 4

Install:

$ npm install open

Usage:

const open = require('open');
 
(async () => {
    // Opens the image in the default image viewer and waits for the opened app to quit.
    await open('unicorn.png', {wait: true});
    console.log('The image viewer app quit');
 
    // Opens the URL in the default browser.
    await open('https://sindresorhus.com');
 
    // Opens the URL in a specified browser.
    await open('https://sindresorhus.com', {app: 'firefox'});
 
    // Specify app arguments.
    await open('https://sindresorhus.com', {app: ['google chrome', '--incognito']});
})();

Answer 5

Windows + Express

app.listen(3000, ()=>{
    require('child_process').exec('start http://localhost:3000/');
});

Answer 6

You may need to implement a switch using the value of ...

require('os').type()

And then use spawn("open") or spawn("xdg-open") depending on the platform?

Answer 7

Simply Use

require('child_process').exec('start https://www.google.co.in/');

It's Worked For me.

Answer 8

The easiest and neatest way, IMHO is using an npm package called openurl. Do a npm install openurl . You could try this real quick in your Nodejs REPL

require("openurl").open("http://stackoverflow.com/questions/8500326/how-to-use-nodejs-to-open-default-browser-and-navigate-to-a-specific-url")

You could also send emails with it if the need arises like so; require("openurl").open("mailto:janedoe@example.com")

Answer 9

#!/usr/bin/env node

const url = 'http://localhost'
require('child_process')
  .exec((process.platform
         .replace('darwin','')
         .replace(/win32|linux/,'xdg-') + 'open ' + url));

Answer 10

var url = '\\index.html';
var start = (process.platform == 'darwin'? 'open': process.platform == 'win32'? 'start': 'xdg-open');
require('child_process').exec(start + ' ' + __dirname + url);

Lexa-B’s answer worked best for me, but I was getting “Windows could not locate index.html” error. I was using lexa-b code child_process exec command to open a local webpage within an npm package I was writing. Needed to open an html file in my npm package when running it / opening it from package bin.js with npx command.

All that was needed was to append __dirname to file path to ensure the relative directory path to the file calling the child_process was correct. The child_process was running at the home folder, which is far away from the npx temp file location. __dirname solves that problem and links the two, solving my missing file error.