忽略gulp.js中的任务依赖项

Question

I have the following task

gulp.task('sass-build', ['clean-build'], function () {
    // do stuff
});

which obviously requires clean-build to be finished before starting. However, I would also like to use this task with gulp.watch without the clean-build dependency.

I'm aware that you could solve this with a helper task like suggested in Skipping Tasks Listed as Dependencies in Gulp , but this would require using gulp.run (which is deprecated) or gulp.start (which isn't even documented because it's considered bad practise).

What would be a better way of doing this? Or is the only way to use gulp.run or gulp.start ?

Answer 1

The approach showed in the linked question is actually not that bad. Instead of gulp.run you can use simple functions which do the same:

var sassFunction = function() {
    return gulp.src('blahbla')
        .pipe(bla())
}

gulp.task('sass-build', ['clean-build'], function () {
    return sassFunction()
});

gulp.task('sass-dev', function () {
    return sassFunction();
});


gulp.task('watch', function() {
    gulp.watch('your/sass/files/**/*.scss', ['sass-dev'])
});

Your other approach is also not so bad and will be fully incorporated with Gulp 4.

gulp.task('default', gulp.series('clean-build', gulp.parallel('sass-build')));

Something to look out for, since it changes dependency trees completely.

Answer 2

I solved this using run-sequence . This also made the entire multitask order much clearer.

What I did exactly was remove the clean-build dependency from sass-build . So eventually:

gulp.task('sass-build', function () {
    // do stuff
});

gulp.task('build', function () {
    runSequence('clean-build', [..., 'sass-build', ...], 'index-build');
});

gulp.task('watch', function () {
    gulp.watch('src/sass/*', ['sass-build']);
});

Very convenient!