is it possible sass compile without running this command code example
Example 1: linux install sass
sudo npm install -g sass
Example 2: compile scss
var gulp = require('gulp');
var sass = require('gulp-sass');
gulp.task("stylesReturn", function () {
return gulp
.src("sass/**/*.scss")
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest("./css/"));
});
gulp.task('stylesDone', function(done) {
gulp
.src("sass/**/*.scss")
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest("./css/"));
done();
});
gulp.task('stylesPromise', function() {
return new Promise(function(resolve, reject) {
gulp
.src("sass/**/*.scss")
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest("./css/"));
resolve();
});
});
gulp.task('stylesAsync', async function() {
gulp
.src("sass/**/*.scss")
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest("./css/"));
});
gulp.task('styleEmitter', function() {
var e = new EventEmitter();
e.on('runstyle',
function(msg) {
gulp.task("stylesReturn", function () {
return gulp
.src("sass/**/*.scss")
.pipe(sass().on("error", sass.logError))
.pipe(gulp.dest("./css/"));
});
console.log(msg);
setTimeout(() => { e.emit('runstyle', 'My message');
e.emit('finish'); });
return e;
});
gulp.task('default',function() {
gulp.watch('sass/**/*.scss', gulp.series('stylesKubernetes'));
});
gulp --tasks
gulp styles
gulp watch
Dir Structure...
--sass
- style.scss
-- css
package.json
Gulpfile.js