How to setup gulp uglify properly?

Question

JavaScript

How to write a task to minify all scripts separately?
If by default I write

gulp.task('javascript', function(){ 
   return gulp.src([ 'app/pre-js/**/*.js' ])
        .pipe(uglify()) 
        .pipe(gulp.dest('app/js'));
});

then all files are compressed into one whole. How to make all files from compressed separately from each other?

Answer 1

Haven't worked with gulp for 1000 years. Perhaps an example of my old config will help you.
When I needed to transpile scss to css and put the result next to the source, I did this:

gulp.task('styles', function ()
{
    return gulp.src(path.join(paths.templatePath, '*.scss'))
  .pipe(sass(sassConfig).on('error', sass.logError))
  .pipe(postcss(postProcessors))
  .pipe(gulp.dest(function (file)
  {
      return file.base;
  }));
});