How to compile only those files that are used in a gulp project?

Question

Node.js

I have a prompt folder, which contains all the plugins and libraries for any occasion.
This folder is compiled into dist.

Question: how to compile only those plugins or libraries that are used at a particular moment in a gulp project?

I use minification for this folder, and so far this code has been written:

gulp.task('prompt', function () {
    return gulp.src('prompt/**/*.js')
        .pipe(uglifyEs())
        .pipe(gulp.dest('dist/prompt')),
    gulp.src('prompt/**/*.css')
        .pipe(csso())
        .pipe(gulp.dest('dist/prompt')),
    gulp.src(['prompt/**/*.*', '!prompt/**/*.css', '!prompt/**/*.js'])
        .pipe(gulp.dest('dist/prompt'));
});

Answer 1

for any occasion.

well, why is that?
Put only the necessary packages from the same npm into the project and build them.
Use modern javascript to easily import packages.
Use babel to build.
npm install jquery

import $ from 'jquery';

$('.selector');

gulp.task('js', function () {
    return gulp.src('src/app.js')
        .pipe(babel())
        .pipe(terser()) // uglify устарел давно
        .pipe(gulp.dest('dist/js')),
});