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如何在Typescript中将类型转换为接口

[英]How to convert type to interface in Typescript

From a typed Object, is it possible to transform it into an interface in TypeScript?从类型化对象,是否可以将其转换为 TypeScript 中的接口?

I already read this QA here on StackOverflow , and don't think it fits well with the description I give in this Question.我已经在 StackOverflow 上阅读了这个 QA ,并且认为它与我在这个问题中给出的描述不太吻合。

Quick Example, this scenario where type Product is defined as type from a third-party TypeScript library.简单的例子,这种场景type Product被定义为type从第三方打字稿库。

# ProductFragmentOne is used to highligt possibility of composition(unions, etc)
type Product = ProductFragmentOne & { sku: string };

To integrate that product in our own system, it is already possible by extending(union) a type as in the following example:要将该产品集成到我们自己的系统中,可以通过扩展(联合)一个类型,如下例所示:

export type ProductSchema = Product & { 
  name: string;
}

My question is:我的问题是:

  • Is there a way for our ProductSchema to be defined as an interface , instead of using the types approach?有没有办法将我们的ProductSchema定义为interface ,而不是使用类型方法? Or is that even possible?或者这甚至可能吗?
# Example of how the code may look
export interface ProductSchema{ 
  name: string;
  // do the magic to add Product properties here
}

Update: The reason for this approach is purely preference of interfaces over types.更新:这种方法的原因纯粹是偏爱接口而不是类型。 Also to make existing code keep its style, regardless of third party library adopted.也使现有代码保持其风格,无论采用何种第三方库。

Thanks.谢谢。

To solve this problem, I used an answer found on a completely unrelated question: Is it "Possible to extend types in Typescript?"为了解决这个问题,我使用了一个完全不相关的问题的答案:“是否可以在 Typescript 中扩展类型?” . .

In fact, since TypeScript 2.2 -- It is possible for an interface to extend a type .事实上,从 TypeScript 2.2 开始—— interface可以扩展type

There is an extensive thread on the difference between interface and type on this StackOverflow thread: TypeScript: Interfaces vs Types在这个 StackOverflow 线程上有一个关于interfacetype之间区别的广泛线程: TypeScript: Interfaces vs Types

export interface ProductSchema extends Product{ 
  name: string;
}
# Where the Product is a type similar to
type Product = { SKU: string }

The "magic trick" I was looking for was indeed that extends operator(or keyword).我正在寻找的“魔术”确实是extends运算符(或关键字)。 There is an additional example on TypeScript playground based on this same approach as wellTypeScript 操场上还有一个基于相同方法的额外示例

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