按需require()

本文介绍了按需require()的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

说我在./libname中创建一个库，其中包含一个主文件:main.js和多个可选库文件，这些文件有时与主对象a.js和b.js一起使用.

Say I create a library in ./libname which contains one main file: main.js and multiple optional library files which are occasionally used with the main object: a.js and b.js.

我创建具有以下内容的index.js文件:

I create index.js file with the following:

exports.MainClass = require('main.js').MainClass; // shortcut exports.a = require('a'); exports.b = require('b');

现在我可以按如下方式使用该库了:

And now I can use the library as follows:

var lib = require('./libname'); lib.MainClass; lib.a.Something; // Here I need the optional utility object lib.b.SomeOtherThing;

但是，这意味着我总是加载"a.js"和"b.js"，而不是在我真正需要它们时加载.

However, that means, I load 'a.js' and 'b.js' always and not when I really need them.

当然可以用require('./libname/a.js')手动加载可选模块，但是随后我丢失了漂亮的lib.a点符号:)

Sure I can manually load the optional modules with require('./libname/a.js'), but then I lose the pretty lib.a dot-notation :)

是否可以通过某种索引文件实现按需加载?也许某些package.json魔术可以在这里很好地发挥作用?

Is there a way to implement on-demand loading with some kind of index file? Maybe, some package.json magic can play here well?

推荐答案

看起来唯一的方法是使用吸气剂.简而言之，像这样:

Looks like the only way is to use getters. In short, like this:

exports = { MainClass : require('main.js').MainClass, get a(){ return require('./a.js'); }, get b(){ return require('./a.js'); } }