ES6默认和命名导出 [英] ES6 default and named exports

问题描述

我正在尝试了解已命名的出口和默认出口.我有一个看似基本的要求，但我不知道如何设置.

I am trying to understand named and default exports. I have a seemingly basic requirement which I don't understand how to setup.

我希望能够同时导入两者:

I want to be able to import both:

//app.js
import Mod from './my-module'
import { funcA, funcB } from './my-module'

console.log('A', Mod.funcA(), funcA()); // A a a
console.log('B', Mod.funcB(), funcB()); // A a a

当我尝试时，最接近的方法是:

When I try, the closest way of doing this I get to is the following:

//my-module.js
export function funcA() { return 'a'; };
export function funcB() { return 'b'; };

export default {funcA, funcB}

我的麻烦是我不想在默认导出中为每个函数重新编制索引.我只想定义我的函数，然后确保将它们导出，以便可以使用其中任何一种方法.

My trouble is that I don't want to reindex each functions in the default export. I just want to define my functions and then make sure they are exported so I can use them either way.

建议?还是我必须使用 import *作为来自'./my-module'; 的Mod ??

Suggestions? Or do I have to use import * as Mod from './my-module';?

推荐答案

您可以省略默认导出，并使用import作为语法:

You can omit the default export and use import as syntax:

//app.js
import * as Mod from './my-module'
import { funcA, funcB } from './my-module'

console.log('A', Mod.funcA(), funcA()); // A a a
console.log('B', Mod.funcB(), funcB()); // B b b

//my-module.js
export function funcA() { return 'a'; };
export function funcB() { return 'b'; };

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