How to spawn a child process with Node.js

Node.js provides a child_process module that provides the ability to spawn child processes.

Require the module, and get the spawn function from it:

const { spawn } = require('child_process')

then you can call spawn() passing 2 parameters.

The first parameter is the command to run.

The second parameter is an array containing a list of options.

Here’s an example:

spawn('ls', ['-lh', 'test'])

In this case you run the ls command with 2 options: -lh and test. This results in the command ls -lh test, which (given that the test file exists in the same folder you run this command in), results in the details about the file:

-rw-r--r--  1 flaviocopes  staff     6B Sep 25 09:57 test

The result of the spawn() function call is an instance of the ChildProcess class that identifies the spawned child process.

Here’s a slightly more complicated example, fully working. We watch the test file and whenever it’s changed, we run the ls -lh command on it:

'use strict'

const fs = require('fs')
const { spawn } = require('child_process')
const filename = 'test'

fs.watch(filename, () => {
  const ls = spawn('ls', ['-lh', filename])
})

There’s one thing missing. We must pipe the output from the child process to the main process, otherwise we won’t see any output from it.

We do so by calling the pipe() method on the stdout property of the child process:

'use strict'

const fs = require('fs')
const { spawn } = require('child_process')
const filename = 'test'

fs.watch(filename, () => {
  const ls = spawn('ls', ['-lh', filename])
  ls.stdout.pipe(process.stdout)
})

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