gulp-strip-import-export

A gulp plugin that removes ES6-style import and export statements

Install

$ npm install --save-dev gulp-strip-import-export

Usage

In your gulpfile.js:

const typescript = require('gulp-typescript');
const stripImportExport = require('gulp-strip-import-export');
 
const tsproject = typescript.createProject({
    module: 'es6',
    noImplicitAny: true,
    target: 'es2018'
});
 
gulp.task('build', function () {
  return gulp.src(['src/*.ts'])
    .pipe(tsproject()).js
    .pipe(stripImportExport())
    .pipe(gulp.dest('dist/js'));
});

Sourcemaps are supported if you've initialized the sourcemaps module.

const sourcemaps = require('gulp-sourcemaps');
 
gulp.task('build', function () {
  return gulp.src(['src/*.ts'])
    .pipe(sourcemaps.init())
    .pipe(tsproject()).js
    .pipe(stripImportExport())
    .pipe(sourcemaps.write())
    .pipe(gulp.dest('dist/js'));
});

What does it do?

Given an example source:

import { Foo } from './foo';
import { Bar } from './bar';
 
export class Baz {
  constructor() { this.foo = new Foo(); }
}

This plugin produces a file stripped of import and export statements:

class Baz {
  constructor() { this.foo = new Foo(); }
}

Why?

This plugin is a corner-case plugin designed for one specific use case: you're using Typescript or Babel to process incoming javascript modules written using ES6-style import and export statements, and you aren't going to wrap your code with a module system (be it webpack, requirejs, rollup, etc.). This allows you to produce your final, browser-ready javascript file by simply concatenating all of the output files in the appropriate order.

This is not a generally useful plugin, but may be useful specifically for projects where you want to minimize the total bytes in the source file -- like js13kgames.