在 ES 6 模块中重新导出默认值 [英] Re-export default in ES 6 modules

问题描述

在 ES6 中，是否可以缩短以下代码.我有一个 App.js 文件和一个 index.js.

In ES6, is it possible to shorten the following code. I have an App.js file and an index.js.

index.js

import App from './App';

export default App;

类似的东西

index.js

export default App from './App.js'

推荐答案

如果你使用 proposal-export-default-from Babel 插件(它是 stage-1 preset)，您将能够使用以下代码重新导出默认值:

If you use proposal-export-default-from Babel plugin (which is a part of stage-1 preset), you'll be able to re-export default using the following code:

export default from "./App.js"

有关更多信息，请参阅ECMAScript 提案.

For more information see the ECMAScript proposal.

另一种方式(没有这个插件)是:

Another way (without this plugin) is:

export { default as App } from "./App.js"

---

当单独的文件(每个文件都有自己的export)具有所有共同点(例如，utils)时，上述是一种非常常见的做法，因此如果，例如, 一个人会想要导入 3 个 utility 函数，而不必编写多个 imports:

---

The above is a very common practice when separate files, each with its own export, have all something in common, for example, utils, so if, for example, one would want to import 3 utility functions, instead of having to write multiple imports:

import util_a from 'utils/util_a' 
import util_b from 'utils/util_b' 
import util_c from 'utils/util_c' 

可以在一行中导入任何实用程序:

One could import any of the utilities in a single-line:

import { util_a, util_b , util_c } from 'utils' 

通过在 /utils 文件夹中创建一个 index.js 文件并导入那里所有实用程序的所有默认值并重新导出，因此 index 文件将作为网关"对于与该文件夹相关的所有导入.

By creating an index.js file in the /utils folder and import all the defaults of all the utilities there and re-export, so the index file will serve as the "gateway" for all imports related to that folder.

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