How can Gulp be restarted upon each Gulpfile chang

Question

I am developing a Gulpfile. Can it be made to restart as soon as it changes? I am developing it in CoffeeScript. Can Gulp watch Gulpfile.coffee in order to restart when changes are saved?

Answer 1

A good solution for Windows, which also works well with Visual Studio task runner.

/// <binding ProjectOpened='auto-watchdog' />
const spawn = require('child-proc').spawn,
      configPaths = ['Gulpconfig.js', 'bundleconfig.js'];

gulp.task('watchdog', function () {
    // TODO: add other watches here

    gulp.watch(configPaths, function () {
        process.exit(0);
    });
});

gulp.task('auto-watchdog', function () {
    let p = null;

    gulp.watch(configPaths, spawnChildren);
    spawnChildren();

    function spawnChildren() {
        const args = ['watchdog', '--color'];

        // kill previous spawned process
        if (p) {
            // You might want to trigger a build as well
            args.unshift('build');

            setTimeout(function () {
                p.kill();
            }, 1000);
        }

        // `spawn` a child `gulp` process linked to the parent `stdio`
        p = spawn('gulp', args, { stdio: 'inherit' });
    }
});

Main changes compared to other answers:

• Uses child-proc because child_process fails on Windows.
• The watchdog exits itself on changes of files because in Windows the gulp call is wrapped in a batch script. Killing the batch script wouldn't kill gulp itself causing multiple watches to be spawned over time.
• Build on change: Usually a gulpfile change also warrants rebuilding the project.