API docs missing gulp.start()

[dman777]

Unless I am having a off day and missing it, it seems gulp.start() is missing from the api docs.

[pkozlowski-opensource]

I think that the start method is on the gulp namespace a bit "by accident" as in reality this is part of the Orchestrator API and gulp inherits from Orchestrator.

I don't think exposing start as public API is what we want to do as the run method is going away in Gulp4 with a new task system.

In short: I don't think it is intended to be Gulp's public API and thus I don't think it should be documented.

[dman777]

Thank you.

What would be an alternative/more proper way to write this to call it from command line? example: gulp foobar

gulp.task('foobar', function () {
    //index is a separate gulp task
    streamReturned = gulp.start('index');
    //do more stuff with `strestreamReturned`;
});

[pkozlowski-opensource]

I would extract the common part of the foobar and index tasks and invoke this function from both tasks. Gulp files is, after all, just a node.js program. Gulp4 will, hopefully, make it more clear that a task is just a function.

At any rate, your problem sounds like a general support question and as such would be better ask on StackOverflow.

[dman777]

👍

[justinmchase]

Start seems like a good api to use in watch tasks. Is there a better way to do this?

gulp.task('watch', () => {
  var sources = [...]
  gulp.watch(sources, ['test'])
  gulp.start('test')
})

If you put test as a dependency instead of invoke via gulp.start then the process will exit if a test fails and watch is never invoked. But this way you will get your tests running immediately but watch will continue to run. This seems like a legitimate enough scenario to add this function to the docs, I had to search for a while to find out how to do this satisfactorily.

[phated]

@justinmchase start is the only way to do what you want in gulp 3 but it was never officially supported by gulp (it was exposed by a dependency) and is completely gone in gulp 4. I don't suggest using it.

[justinmchase]

how would you do it in gulp 4, for reference?

[phated]

@justinmchase everything in gulp 4 is just a function, you just call it

[justinmchase]

I see, thanks.

[felixfbecker]

everything in gulp 4 is just a function, you just call it

@phated Is there a way to make Gulp log the standard Starting 'taskName'/Finished 'taskName' after XXms when doing so?

[demurgos]

Is your use case related to the watch mode or is it about simply starting the tasks programmatically? There was a recent issue for this to trigger gulp tasks programmatically that got shut down (simply call the functions, as mentioned above).

The logs are handled at the CLI level. I don't think there's way to trigger them when calling the task function programmatically.

I'd recommend to open a new issue if you want to request this feature instead of replying to an old issue. If this is just a support question, it's better suited for Stack Overflow. (Due to the high number of issues, this tracker is only intended for bugs)

[felixfbecker]

It's a watch use case, but when not using gulp.watch(), but e.g. webpack's watch API. After the webpack compilation is done I also need to trigger some other tasks.

[demurgos]

Seems like a valid use case to me but I am not sure if it is possible currently.

[phated]

Shell out or refactor your project. This is not a support forum so I'm going to lock this.