title: Gulp Sass - How to properly name the output css?
body: <p>I'm reading a tutorial about sass <a href="https://scotch.io/tutorials/getting-started-with-sass" rel="noreferrer">here</a> 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</p>
  
  <pre><code>gulp.task('compileNavbar', function() {
      gulp.src('assets/css/sass/**/*.navbar.scss')
          .pipe(sass('navbar.css'))
          .pipe(gulp.dest('assets/css/'));;
  });
  </code></pre>
  
  <p>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 <strong>BUT</strong> the css is named as <strong>guest.navbar.css</strong> which I didnt expect. I want it to be named as <strong>navbar.css</strong> but how to?</p>