Make two files instead of three

issue に示されている factor-bundle コマンドから始め、lib.js、main.js、checkout.js を使って再現します。コマンドが共有バンドルとエントリーバンドルをどのように処理するかを追跡します。完了の条件は、./main を必要とせず、junk.js も生成せずに、要求されたサイト用バンドルと checkout バンドルを生成することです。

Hey!

I effectively want to split out part of a file, instead of splitting out the common parts of two files.

Currently I have bundle.js, which contains everything used on the site. There is some checkout specific logic there that I only want on the checkout, so I want to split that stuff out into bundle-checkout.js, but then run both bundle.js and bundle-checkout.js in the checkout.

With factor-bundle by default, I would be running these files on the main site:

• bundle-common.js
• bundle.js

And these on the checkout:

• bundle-common.js
• bundle.js
• bundle-checkout.js

It's pointless for me to have both bundle-common.js and bundle.js!

The best solution I've found so far is to add require('./main') to the top of the input bundle-checkout.js, which would mean that bundle.js is effectively an empty file and I can include bundle-common.js by itself, but then that generates a junk file.

lib.js:

module.exports = 'lib';

main.js:

var lib = require('./lib');
console.log(lib);

checkout.js:

require('./main');
console.log('special checkout logic');

And then the command:

browserify main.js checkout.js -p [ factor-bundle -o junk.js -o bundle-checkout.js ] -o bundle.js

Is there any way to do this without the require('./main') and the useless junk.js file?

Thanks, and sorry for the long question! I'll turn this into an article once it's been figured out so that nobody asks it again :)