gulp notify two tasks when watch invoked task completes - gulp

Question

I have a gulpfile in which is a watch task, below
gulp.task('watch', function() {
gulp.watch('template/slick/assets/less/*.less', ['less']); // Watch all the .less files, then run the less task
});
This then invokes a less compile to css of that directory and moves it to a /css/ folder.
I then have 2 other tasks already scripted up to min, concat and move these .css files to a dist folder.
What I need to know is that when my watch invoked less task completes can I notify/run the stylesmin, cssconcats tasks? I do need to add more code to do it. I can't see to find a decent notify/end style way of doing things.
Here's the less task which is invoked by watch
gulp.task('less', function () {
return gulp.src('game/http/template/slick/assets/less/*.less')
.pipe(less())
.pipe(gulp.dest('game/http/template/slick/assets/css/'))
.pipe(notify({message: 'Less compiled'}));
});

Answer 1

You can define your tasks as dependencies of each other with [] syntax like below. (I'm assuming you are using something like connect to start up a server before you start watching the files)
gulp.task('less', function() {
console.log('less');
})
gulp.task('stylesmin', ['less'], function() {
console.log('stylesmin');
})
gulp.task('cssconcats', ['stylesmin'], function() {
console.log('cssconcats');
})
gulp.task('test', ['connect', 'watch']);
Modify your watch to kick off the last task, cssconcats and gulp will run the dependencies first
gulp.task('watch', function() {
gulp.watch('template/slick/assets/less/*.less', ['cssconcats']); // Watch all the .less files, then run the less task
});
Change any .less file and the output shows that the tasks are run correctly in this order;
less, stylesmin, cssconcats