如何在 TypeScript 模块中拥有具有多种类型和值的“导出默认值”？

Question

我可以轻松地获得具有多个值的导出默认值：

class Car {...}
class Bus {...}

export default { Car, Bus }

我还可以轻松获得某种类型的导出默认值

export default interface Airplane {...}

但我不能默认导出多种类型

interface Airplane {...}
interface Motorcycle {...}

// 'Airplane' only refers to a type, but is being used as a value here. ts(2693)
export default { Airplane, Motorcycle }

或者多种类型和值的混合。

class Car {...}
interface Motorcycle {...}

// 'Airplane' only refers to a type, but is being used as a value here. ts(2693)
export default { Car, Airplane }

我怎样才能实现这个目标？

Answer 1

考虑export =语法。

该export =语法指定从模块导出的单个对象。这可以是类、接口、命名空间、函数或枚举。

车辆.ts

namespace Vehicle {
    export class Car {
        // ...
    }

    export class Bus {
        // ...
    }
}

export = Vehicle;

测试1.ts

import Vehicle from './Vehicle';

const car = new Vehicle.Car();
const bus = new Vehicle.Bus();

测试2.ts

import { Bus, Car } from './Vehicle';

const car = new Car();
const bus = new Bus();

请参阅https://www.typescriptlang.org/docs/handbook/modules.html#export--and-import--require

Answer 2

事实上，通过像这样导出：

class Car {...}
interface Motorcycle {...}

// 'Airplane' only refers to a type, but is being used as a value here. ts(2693)
export default { Car, Airplane }

您要做的就是**导出具有两个属性的默认对象*。*

稍后，在您的代码中，您可以像这样导入它：

// name is as you like, its default import
import transport from transport;

并像这样使用它：

....new transport.Car()
....new transport.Plane()