如何以特定顺序运行Gulp任务

Question

I am using runsequence plugin. But It seems that I am doing something wrong so it is not working. I am trying to create browserify build and concat it with my handlebars precompiled templates on server using run-sequence gulp plugin.

In gulpfile correspoding tasks looks like this:

gulp.task("browserify", function () {
    gulp.src('src/js/map.js')
        .pipe(browserify({
            insertGlobals: false,
            debug: false
        }))
        .pipe(gulp.dest('build'))
        .pipe(notify({ message: "browserify ended" }));
});

gulp.task("scripts", function () {
    var newDir;

    newDir = './build/1111';
    gulp.src(['./build/map.js', './build/templates/templates.js'])
        .pipe(concat('build.js'))
        .pipe(gulp.dest(newDir))
        .pipe(notify({ message: "scripts build ended" }));
});

gulp.task('default', function () {
    runSequence( 'browserify', 'scripts' );
});

But actually scripts task are trying invoke before browserify task is done. So I get this output:

$ gulp
[10:40:26] Using gulpfile ~/projects/map2/gulpfile.js
[10:40:26] Starting 'default'...
[10:40:26] Starting 'browserify'...
[10:40:26] Finished 'browserify' after 7.84 ms
[10:40:26] Starting 'scripts'...
[10:40:26] Finished 'scripts' after 4.58 ms
[10:40:26] Finished 'default' after 14 ms
[10:40:26] gulp-notify: [Gulp notification] templates build ended
[10:40:26] gulp-notify: [Gulp notification] browserify ended

And scripts task don't create build.js file (note there is no message "scripts build ended" from notify plugin).

Answer 1

I figured out it myself. For achieving this there should be some signal from task about finishing work. Check this article for more information. Actually I choose to return stream like this:

gulp.task('browserify', function (cb) {
    var stream = gulp.src('src/js/clientAppControl.js')
        .pipe(browserify({
            insertGlobals: false,
            debug: false
        }))
        .pipe(gulp.dest('build'))
        .pipe(notify({ message: 'browserify ended' }));

    return stream;
});

and dependencies started to work properly.

Answer 2

Similar to the answer you added, but a little cleaner in my opinion, is to remove return stream and replace var stream = with return .

gulp.task('browserify', function (cb) {
    return gulp.src('src/js/clientAppControl.js')
        .pipe(browserify({
            insertGlobals: false,
            debug: false
        }))
        .pipe(gulp.dest('build'))
        .pipe(notify({ message: 'browserify ended' }));
});