By what should I replace gulp.run ?

[edi9999]

Hi I have the following tasks:

gulp.task('watch', function () {
    gulp.src(paths.coffee)
        .pipe(watch(function(files) {
            var f=files.pipe(coffee())
                .pipe(gulp.dest(paths.js))
            gulp.run('jasmine');
            gulp.run('livereload');
            return f;
        }));
    gulp.src(paths.stylus)
        .pipe(watch(function(files){
            var f=gulp.src(paths.mainStylus)
                .pipe(stylus())
            f
                .pipe(gulp.dest(paths.css))
            gulp.run('livereload');
            return f;
        }))
});
gulp.task('livereload',function(cb)
        {
            server.changed('SpecRunner.html');
            cb()
        })

gulp.task('jasmine', function(cb) {
...
});

How should I in that sample, replace the run method which is currently deprecated ?

[sindresorhus]

Questions are better suited for StackOverflow.

But you can use .start instead of .run https://github.com/orchestrator/orchestrator/#orchestratorstarttasks-cb

[robrich]

On your way to stack overflow, try one of these:

• use .start() instead of .run() -- 98% of the time you shouldn't do this
• change it to this: gulp.watch(['globs'], ['array','of','tasks'])

[dman777]

@sindresorhus Do you intend for .start to be used since you mentioned this? it's not in the api of Gulp. The reason being, not sure if using this will cause breakage in Gulp 4.0 and stackflow doesn't have any answers to this(but many of the same question).

[yocontra]

This is an issue of bad code organization, not something to do with .start().

You want something like this:

function reload() {
  server.changed('SpecRunner.html');
}

gulp.task('watch', function() {
  gulp.src(paths.coffee)
    .pipe(watch(function(files) {
      return files.pipe(coffee())
        .pipe(gulp.dest(paths.js))
        .once('end', reload);
    }));

  gulp.src(paths.stylus)
    .pipe(watch(function(files) {
      return gulp.src(paths.mainStylus)
        .pipe(stylus())
        .pipe(gulp.dest(paths.css))
        .once('end', reload);
    }))
});

[dman777]

I see gulp.start() being used in production environments and all over stackoverflow, and it would be nice to have some clarification on start and if the
https://medium.com/@contrahacks/gulp-3828e8126466

Orchestrator is undergoing a huge overhaul

is going to make start deprecated? as far as Orchestrator being overhauled, there is nothing listed in it's milestone page on github. At this point, clarification on start is important to plan for breakage when 4.0 comes out.

[yocontra]

@dman777 Here is my answer: don't use .start, you don't need it. It's an internal function used by the CLI. It doesn't need to be deprecated because it is already undocumented and marked as internal. It has been since version 1.0 almost a year ago. Named functions work perfectly fine. There is no reason to make something a task unless you intend to call it from the CLI. Even if you need to call it from the CLI, you can still use named functions.

Repeat after me:

Use functions. You don't need .start().
Use functions. You don't need .start().
Use functions. You don't need .start().

[phated]

The 4.0 branch already has start removed. Did i mention you should start using the 4.0 branch and submit issues for us so we can ship it?

[dman777]

👍