How to start a process in Gulp.js with a shell command, and then kill it?

Question

JavaScript

I would like to run a test local server for testing ReST requests.

// Запустить можно командой:
gulp.task('startTestServ', ['beforeTask'], shell.task([
  './myTestServ &', // команда выведет в консоль id процесса
]));
// После этого запускаем, собственно, тесты:
gulp.task('restTests', ['startTestServ'], function() {
  return gulp.src(['restTests/*.js'], {read: false})
    .pipe(mocha());
});
// В конце убиваем тестовый сервер
gulp.task('killTestServ', shell.task([
  'kill ???', // Но не знаем id процесса
]));

How can I get the id of a running process using gulp-shell or a similar tool?
Of course, you can get around the problem in every possible way, but the question is about getting the id of the running process in gulp.

Answer 1

The standard child_process.spawn is enough:

var gulp = require('gulp');
var spawn = require('child_process').spawn;

var ping;

gulp.task('ping', function() {
    ping = spawn('ping', ['localhost'], {
        stdio: 'inherit' // для примера, серверу не нужно
    });
});

gulp.task('test', ['ping'], function() {
    setTimeout(function() {
        ping.kill();
    }, 2000);
});

Answer 2

Have a look at gulp-run , it makes it easy to capture and write stdout, I think this will solve your problem