Gulp Sass - How to properly name the output css?
问题 I'm reading a tutorial about sass here then I tried some other approach and I cant get the answer in this tutorial. Theres the problem. I have this code in my gulpfile.js gulp.task('compileNavbar', function() { gulp.src('assets/css/sass/**/*.navbar.scss') .pipe(sass('navbar.css')) .pipe(gulp.dest('assets/css/'));; }); As of now, I only have assets/css/sass/guest.navbar.scss and this code locates the scss files correctly and puts the output css file in the correct directory BUT the css is