将 Typescript 接口中的所有属性设为可选 [英] Make all properties within a Typescript interface optional

问题描述

我的应用程序中有一个接口:

I have an interface in my application:

interface Asset {
  id: string;
  internal_id: string;
  usage: number;
}

这是帖子界面的一部分:

that is part of a post interface:

interface Post {
  asset: Asset;
}

我还有一个用于后期草稿的界面，其中资产对象可能只是部分构建

I also have an interface that is for a post draft, where the asset object might only be partially constructed

interface PostDraft {
  asset: Asset;
}

我想允许 PostDraft 对象拥有部分资产对象，同时仍然检查那里的属性的类型(所以我不想只是用 any).

I want to allow a PostDraft object to have a partial asset object while still checking types on the properties that are there (so I don't want to just swap it out with any).

我基本上想要一种能够生成以下内容的方法:

I basically want a way to be able to generate the following:

interface AssetDraft {
  id?: string;
  internal_id?: string;
  usage?: number;
}

无需完全重新定义Asset 接口.有没有办法做到这一点?如果没有，在这种情况下安排我的类型的明智方法是什么?

without entirely re-defining the Asset interface. Is there a way to do this? If not, what would the smart way to arrange my types in this situation be?

推荐答案

这在 TypeScript 中是不可能的 <2.1 无需创建具有可选属性的附加接口；但是，这可以通过在 TypeScript 2.1+ 中使用映射类型来实现.

This isn't possible in TypeScript < 2.1 without creating an additional interface with optional properties; however, this is possible by using mapped types in TypeScript 2.1+.

为此，请使用 TypeScript 默认提供的 Partial 类型.

To do this, use the Partial<T> type which TypeScript provides by default.

interface PostDraft {
    asset: Partial<Asset>;
}

现在 asset 上的所有属性都是可选的，这将允许您执行以下操作:

Now all the properties on asset are optional, which will allow you to do the following:

const postDraft: PostDraft = {
    asset: {
        id: "some-id"
    }
};

关于部分

部分 是定义 为映射类型，使所提供类型中的每个属性都可选(使用 ? 标记).

Partial<T> is defined as a mapped type that makes every property in the provided type optional (using the ? token).

type Partial<T> = {
    [P in keyof T]?: T[P];
};

阅读有关映射类型的更多信息此处 和在手册中.

Read more about mapped types here and in the handbook.

深度局部

如果您想要在对象上递归工作的部分实现，那么您可以在 TS 4.1+ 中使用以下类型:

If you want a partial implementation that works recursively on objects then you can use the following type in TS 4.1+:

type DeepPartial<T> = {
    [P in keyof T]?: T[P] extends object ? DeepPartial<T[P]> : T[P];
};

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