Is it possible to use variables in gulpfile.js?

Question

JavaScript

I want to set the name of the template, for example: $mytemplate
And then I need to substitute this value in the tasks.

gulp.task('sass', function(){
  return gulp.src('templates/$mytemplate/asset/sass/*.scss')
    .pipe(sass())
    .pipe(gulp.dest('templates/$mytemplate/css/'))
    .pipe(browserSync.stream());
});

Answer 1

You can, because it's just regular JS.
You are apparently a beginner PHP developer, I am deriving it from single quotes and a variable in a string.