gulp.run is deprecated. How do I compose tasks?

gulp, javascript

Solution

gulp.task('watch', function () {
  var server = ['jasmine', 'embed'];
  var client = ['scripts', 'styles', 'copy', 'lint'];
  gulp.watch('app/*.js', server);
  gulp.watch('spec/nodejs/*.js', server);
  gulp.watch('app/backend/*.js', server);
  gulp.watch('src/admin/*.js', client);
  gulp.watch('src/admin/*.css', client);
  gulp.watch('src/geojson-index.json', ['copygeojson']);
});

You no longer need to pass a function (though you still can) to run tasks. You can give watch an array of task names and it will do this for you.

Problem

Here is a composed task I don't know how to replace it with task dependencies. ``` ... gulp.task('watch', function () { var server = function(){ gulp.run('jasmine'); gulp.run('embed'); }; var client = function(){ gulp.run('scripts'); gulp.run('styles'); gulp.run('copy'); gulp.run('lint'); }; gulp.watch('app/*.js', server); gulp.watch('spec/nodejs/*.js', server); gulp.watch('app/backend/*.js', server); gulp.watch('src/admin/*.js', client); gulp.watch('src/admin/*.css', client); gulp.watch('src/geojson-index.json', function(){ gulp.run('copygeojson'); }); }); ``` The corresponding changelog https://github.com/gulpjs/gulp/blob/master/CHANGELOG.md#35 [deprecate gulp.run]

Original source