How to run Gulp tasks sequentially one after the other

Question

in the snippet like this:

gulp.task \"coffee\", ->
    gulp.src(\"src/server/**/*.coffee\")
        .pipe(coffee {bare: true}).on(\"error\",gutil.log)
           


        

Answer 1

Gulp and Node use promises.

So you can do this:

// ... require gulp, del, etc

function cleanTask() {
  return del('./dist/');
}

function bundleVendorsTask() {
  return gulp.src([...])
    .pipe(...)
    .pipe(gulp.dest('...'));
}

function bundleAppTask() {
  return gulp.src([...])
    .pipe(...)
    .pipe(gulp.dest('...'));
}

function tarTask() {
  return gulp.src([...])
    .pipe(...)
    .pipe(gulp.dest('...'));
}

gulp.task('deploy', function deployTask() {
  // 1. Run the clean task
  cleanTask().then(function () {
    // 2. Clean is complete. Now run two tasks in parallel
    Promise.all([
      bundleVendorsTask(),
      bundleAppTask()
    ]).then(function () {
      // 3. Two tasks are complete, now run the final task.
      tarTask();
    });
  });
});

If you return the gulp stream, you can use the then() method to add a callback. Alternately, you can use Node's native Promise to create your own promises. Here I use Promise.all() to have one callback that fires when all the promises resolve.