How to fork a process of another module

Question

TL;DR : How does one fork a process that is located outside of the current running process?

I'm trying to use child_process of Nodejs in order to start another nodejs process on the parent's process exit.

I successfully executed the process with the exec but I need the child process be independent of the parent, so the parent can exit without waiting for the child, hence I tried using spawn with the detached: true, stdio: 'ignore' option and unref()ing the process:

setting options.detached to true makes it possible for the child process to continue running after the parent exits.

spawn('node MY_PATH', [], {detached: true, stdio: 'ignore'}).unref();

This yields the :

node MY_PATH ENOENT error. which unfortunately I've failed resolve.

After having troubles achieving this with spawn and reading the documentationagain i figured i should actually use fork:

The child_process.fork() method is a special case of child_process.spawn() used specifically to spawn new Node.js processes.

fork() doesnt take a command as its' first argument, but a modulePath which i can't seem to fit since the script I'm trying to run as a child process isnt in the directory of the current running process, but in a dependency of his.

Back to the starting TL;DR - how does one fork a process that is located outside of the current running process?

Any help would be much appreciated!

EDIT:

Providing a solution to the spawn ENOENT error could be very helpfull too!

Answer 1

Following code should allow you to do what you need.

var Path = require('path');
var Spawn = require('child_process').spawn;

var relative_filename = '../../node_modules/bla/bla/bla.js';

Spawn('node', [Path.resolve(__dirname, relative_filename)], {detached: true, stdio: 'ignore'}).unref();

process.exit(0);