gulp minify all css files to a single file

Question

I have gulp task script as following,

// loads various gulp modules
var gulp = require(\'gulp\');
var concat = require(\'gulp-concat\');
var minifyCSS = requ         


        

Answer 1

Your gulp task is missing the concat pipe.

gulp.src('src/css/**/*.css')
    .pipe(minifyCSS())
    .pipe(autoprefixer('last 2 version', 'safari 5', 'ie 8', 'ie 9'))
    .pipe(concat('style.min.css'))
    .pipe(gulp.dest('dist/css'))

Here is a very good tutorial about building with gulp:

http://www.smashingmagazine.com/2014/06/11/building-with-gulp/

Answer 2

I think better is used the cleancss gulp plugin.

const cleanCSS = require('gulp-clean-css');  // npm install gulp-clean-css --save-dev

gulp.task('css', () => {

return gulp.src('assets/styles/*.css')
  .pipe(cleanCSS({
      debug: true,
      compatibility: 'ie8',
      level: {
          1: {
              specialComments: 0,
          },
      },
  }))
  .pipe(autoprefixer({
      browsers: ['last 2 versions'],
      cascade: false
  }))
  .pipe(rename({
      basename: 'main-styles',
      suffix: '.min',
  }))
  .pipe(gulp.dest('dist/assets/styles/'))

});