Ignoring SCSS/SASS partials in Gulp task -

i've got following structure:

├── assets │   └── css │       └── output.css ├── gulpfile.js ├── index.html ├── package.json └── scss     ├── some-other.css     ├── _base.scss     ├── _config.scss     ├── _helpers.scss     ├── _home.scss     └── main.scss 

i'm using gulp compile css , scss files in scss/ 1 file in assets/css/. relevant task gulpfile.js:

gulp.task('build-css', function() {   return gulp.src(['scss/**/*.scss','scss/**/*.css'])     .pipe(concat('output.css'))     .pipe(sass())     .pipe(gulp.dest('assets/css')); }); 

the problem scss/sass partials (those files starting underscore in scss/ fed through the task , fail.

i'd ignore files anywhere in scss/ start underscore , end .scss.

i've taken @ following wasn't able work:

• gulp doesn't ignore sass partials underscore
• how can use glob ignore files start underscore?

it seems mislead , using suggestions in linked questions able working following:

gulp.task('build-css', function() {   return gulp.src('scss/**/[^_]*.?(s)css')     .pipe(concat('output.css'))     .pipe(sass())     .pipe(gulp.dest('assets/css')); }); 

sorry noise - should have been more thorough in testing before asking question. hope helps someone, though.