I have two sets of files, let's call them base and mods. The mods files override the base files, so when I run the gulp task related to base, I need to run the mods task directly after. My setup is something like this:

gulp.task('base',function(){
  return gulp.src('base-glob')
    .pipe(...)
    .pipe(gulp.dest('out-glob'))
});

gulp.task('mods',function(){
  return gulp.src('mods-glob')
    .pipe(...)
    .pipe(gulp.dest('out-glob'))
});

So I want to run the mods task at the pletion of the base task. Note that this is not the same as defining base as a dependency of mods, because if I'm only changing mods files, I only need to run the mods task. I'd prefer not to use a plugin.

I've been reading the docs about callback functions and other suggestions of synchronous tasks, but can't seem to get my head around it.

I have two sets of files, let's call them base and mods. The mods files override the base files, so when I run the gulp task related to base, I need to run the mods task directly after. My setup is something like this:

gulp.task('base',function(){
  return gulp.src('base-glob')
    .pipe(...)
    .pipe(gulp.dest('out-glob'))
});

gulp.task('mods',function(){
  return gulp.src('mods-glob')
    .pipe(...)
    .pipe(gulp.dest('out-glob'))
});

So I want to run the mods task at the pletion of the base task. Note that this is not the same as defining base as a dependency of mods, because if I'm only changing mods files, I only need to run the mods task. I'd prefer not to use a plugin.

I've been reading the docs about callback functions and other suggestions of synchronous tasks, but can't seem to get my head around it.

• javascript
• gulp
• synchronous

• You have conflicting requirements. You want mods to run AFTER base but you only want mods to run by itself if only mods files are changed? That doesn't make sense exactly. – Jake Wilson Commented Feb 9, 2015 at 19:13
• Sorry I wasn't clear. I have two use cases: when base files change, I want to run base then mods. When mods files change I only want to run mods. – nvioli Commented Feb 9, 2015 at 19:17
• I figured as much just wanted to make sure. See my answer below. – Jake Wilson Commented Feb 9, 2015 at 19:20
• possible duplicate of How to run Gulp tasks synchronously/one after the other – Andrew Marshall Commented Jun 15, 2015 at 17:50

2 Answers 2

I know you don't want to use a plugin, but gulp doesn't have a way to run a sequence of tasks in order without a plugin. Gulp 4 will, but in the meantime the stopgap solution is the run-sequence plugin.

gulp.task('all', function() {
  runSequence('base', 'mods');
});

This ensures that the tasks run in order as opposed to unordered dependencies.

Now setup a watch:

gulp.task('watch', function() {
    gulp.watch('base-glob', ['all']);
    gulp.watch('mods-glob', ['mods']);
});

Whenever base-glob changes, gulp will run all task, which will run the sequence base then mods.

Whenever mods-glob changes, gulp will run only mods task.

That sound about right?

runSequence has some weird bugs, it kept plaining that my tasks are not defined.

If you look into the Orchestrator source, particularly the .start() implementation you will see that if the last parameter is a function it will treat it as a callback.

I wrote this snippet for my own tasks:

  gulp.task( 'task1', () => console.log(a) )
  gulp.task( 'task2', () => console.log(a) )
  gulp.task( 'task3', () => console.log(a) )
  gulp.task( 'task4', () => console.log(a) )
  gulp.task( 'task5', () => console.log(a) )

  function runSequential( tasks ) {
    if( !tasks || tasks.length <= 0 ) return;

    const task = tasks[0];
    gulp.start( task, () => {
        console.log( `${task} finished` );
        runSequential( tasks.slice(1) );
    } );
  }
  gulp.task( "run-all", () => runSequential([ "task1", "task2", "task3", "task4", "task5" ));