How to write task correctly?

Question

gulp.js

gulp.task('css', function() {
    return gulp.src([
            'node_modules/normalize.css/normalize.css',
            'node_modules/slick-carousel/slick/slick.css'
        ])
        .pipe(concat('_libs.scss'))
        .pipe(gulp.dest('app/scss'))
        .pipe(browserSync.reload({ stream: true }))
});

Can you tell me how to change this task without concatenation, if I need one normalize.css file without slick-carousel and so that it is also written to the _libs.scss file

Answer 1

gulp.task('css', function() {
    return gulp.src([
            'node_modules/normalize.css/normalize.css'
        ])
        .pipe(concat('_libs.scss'))
        .pipe(gulp.dest('app/scss'))
        .pipe(browserSync.reload({ stream: true }))
});

As an option like this) Probably not the most elegant solution, but you don’t need to change anything except 1 line, but it will work