There are several tasks defined with gulp.task in gulpfile.js:
gulp.task('sync-task', () => { ... });
gulp.task('async-task', cb => { ... });
I would like to start one of them programmatically. Preferably in the same process (no exec, etc.), because one of the reasons behind this is the ability to run the script in debugger.
How can this be done?
It looks like there were things like gulp.run and gulp.start, but they are deprecated in Gulp 4.
It appears that a task can be retrieved with gulp.task getter function and called directly:
gulp.task('sync-task')();
gulp.task('async-or-random-task')(function callbackForAsyncTasks(err) {
if (err)
console.error('error', err);
});
gulp.task('sync-task', () => {...}).
[12:00:00] Starting 'sync-task'... and [12:00:01] Finished 'sync-task' after 1000 ms to be outputted, but I'm not getting that. Is this still possible, if yes, how?Adding to the accepted answer:
Yes you can retrieve a task by calling gulp.task, but rather than calling it directly you can wrap it in a call to gulp.series (presumably gulp.parallel would also work) and call that. This has the benefit of producing the expected "Starting/Finished" output. An example would be:
gulp.series(gulp.task('some-task'))();
with the output:
[09:18:32] Starting 'some-task'...
[09:18:32] Finished 'some-task' after 7.96 ms
This was tested just now on gulp v4.0.2.
I can run a task by doing this:
gulp.task('sync-task').unwrap()();
the gulp.task('sync-task').unwrap() returns the original task function But it is not called as Gulp task function (synchronous call only), the easiest way to grammatically call Gulp tasks is to use the parallel(task1, ...)() or series(task1, ...)() methods (from Gulp: v4.0.0 the task() method is discouraged to use (but supported))