Piping assets to a destination without modification

I’m just getting started with streams, and I’m hitting a bit of a wall. Basically, I want to get the assets picked up in my index.html build blocks, and write them to a dev folder, preserving their src attributes. But for some reason, this doesn’t output anything:

index.html

<!-- build:js(./bower_components/) libraries.js -->
<script src="lodash/lodash.js"></script>
<script src="firebase/firebase.js"></script>
<!-- endbuild -->

Gulpfile.js

gulp.task('pipeAssets', function () {
    return gulp.src('app/*.html')
        .pipe(useref.assets({ noconcat: true }))
        .pipe(debug())
        .pipe(gulp.dest('./output'));
});

I just want to do a straight copy of the files - and they’re in the stream if I use debug() - where am I going wrong? Thanks 😃

[palmerj]

Having a little tap function like @amaeland also worked for me:

// Create output HTML pages from templates then
// concat & minify scripts and css
gulp.task('html', ['jshint', 'sass'], () => {
    return gulp.src('src/html/pages/**/*.hbs')
        .pipe(plumber())
        .pipe(handlebars({}, {
            ignorePartials: true,
            batch: ['src/html/partials']
        }))
        .pipe(rename({
            extname: '.html'
        }))
        .pipe(useref({ noconcat: development() }))
        .pipe(tap(function(file, t) {
            // ensure asset base path is aligned with
            // useref block asset paths within the html files
            if (['.js', '.css'].indexOf(file.extname) >= 0) {
                file.base = 'src';
            }
            return file;
        }))
        .pipe(gulpif('*.js', production(uglify())))
        .pipe(gulpif('*.css', production(cleanCss())))
        .pipe(gulp.dest('dist/'))
        .pipe(browserSync.stream());
});

[joshbernfeld]

Thank you @palmerj