Execute bash command in Node.js and get exit code

Question

I can run a bash command in node.js like so:

var sys = require(\'sys\')
var exec = require(\'child_process\').exec;

function puts(error, stdout, stderr) { s         


        

Answer 1

child_process.spawnSync()

This function exposes the nicest sync interface: https://nodejs.org/api/child_process.html#child_process_child_process_spawnsync_command_args_options

Example:

#!/usr/bin/env node

const child_process = require('child_process');
let out;

out = child_process.spawnSync('true');
console.log('status: ' + out.status);
console.log('stdout: ' + out.stdout.toString('utf8'));
console.log('stderr: ' + out.stderr.toString('utf8'));
console.log();

out = child_process.spawnSync('false');
console.log('status: ' + out.status);
console.log('stdout: ' + out.stdout.toString('utf8'));
console.log('stderr: ' + out.stderr.toString('utf8'));
console.log();

out = child_process.spawnSync('echo', ['abc']);
console.log('status: ' + out.status);
console.log('stdout: ' + out.stdout.toString('utf8'));
console.log('stderr: ' + out.stderr.toString('utf8'));
console.log();

Output:

status: 0
stdout: 
stderr: 

status: 1
stdout: 
stderr: 

status: 0
stdout: abc

stderr: 

Tested in Node.js v10.15.1, Ubuntu 19.10.