Gulp: target to debug mocha tests

Question

I have a set of gulp.js targets for running my mocha tests that work like a charm running through gulp-mocha . Question: how do I debug my mocha tests running through gulp? I would like to use something like node-inspector to set break points in my src and test files to see what's going on. I am already able to accomplish this by calling node directly:

node --debug-brk node_modules/gulp/bin/gulp.js test

But I'd prefer a gulp target that wraps this for me, eg:

gulp.task('test-debug', 'Run unit tests in debug mode', function (cb) {
   // todo?
});

Ideas? I want to avoid a bash script or some other separate file since I'm trying to create a reusable gulpfile with targets that are usable by someone who doesn't know gulp.

Here is my current gulpfile.js

// gulpfile.js
var gulp = require('gulp'),
  mocha = require('gulp-mocha'),
  gutil = require('gulp-util'),
  help = require('gulp-help');

help(gulp); // add help messages to targets

var exitCode = 0;

// kill process on failure
process.on('exit', function () {
  process.nextTick(function () {
    var msg = "gulp '" + gulp.seq + "' failed";
    console.log(gutil.colors.red(msg));
    process.exit(exitCode);
  });
});

function testErrorHandler(err) {
  gutil.beep();
  gutil.log(err.message);
  exitCode = 1;
}

gulp.task('test', 'Run unit tests and exit on failure', function () {
  return gulp.src('./lib/*/test/**/*.js')
    .pipe(mocha({
      reporter: 'dot'
    }))
    .on('error', function (err) {
      testErrorHandler(err);
      process.emit('exit');
    });
});

gulp.task('test-watch', 'Run unit tests', function (cb) {
  return gulp.src('./lib/*/test/**/*.js')
    .pipe(mocha({
      reporter: 'min',
      G: true
    }))
    .on('error', testErrorHandler);
});

gulp.task('watch', 'Watch files and run tests on change', function () {
  gulp.watch('./lib/**/*.js', ['test-watch']);
});

Answer 1

With some guidance from @BrianGlaz I came up with the following task. Ends up being rather simple. Plus it pipes all output to the parent's stdout so I don't have to handle stdout.on manually:

  // Run all unit tests in debug mode
  gulp.task('test-debug', function () {
    var spawn = require('child_process').spawn;
    spawn('node', [
      '--debug-brk',
      path.join(__dirname, 'node_modules/gulp/bin/gulp.js'),
      'test'
    ], { stdio: 'inherit' });
  });

Answer 2

You can use Node's Child Process class to run command line commands from within a node app. In your case I would recommend childprocess.spawn() . It acts as an event emitter so you can subscribe to data to retrieve output from stdout . In terms of using this from within gulp, some work would probably need to be done to return a stream that could be piped to another gulp task.