从可区分联合属性到对象类型的映射 [英] Mapping from discriminated union property to object type

问题描述

我有一堆对象，它们都有一个属性来区分它们.我将它们作为联合类型.现在我想创建一个从被区分属性到实际类型的映射.我可以自己制作，但它具有两面性且容易出错，所以我想知道是否有某种方法可以使用 TypeScript 以编程方式生成它.:)

I have a bunch of objects that all have a property to discriminate them. And I have them as a union type. Now I want to create a mapping from the discriminated property to the actual type. I can make it myself, but its duplicitous and error prone so I was wondering if there was some way of generating this programmatically with TypeScript. :)

type X = { type: "x", x: number }
type Y = { type: "y", y: number }

type Value = X | Y
type Type = Value["type"]

// Is it possible to generate this?
type TypeToValue = {
    x: X,
    y: Y,
}

// Its useful for stuff like this
function getRecord<T extends Type>(type: T, id: string): TypeToValue[T] {
    return null as any
}

推荐答案

如果签名始终与您的帖子相同，而 type 的值必须匹配关联属性(例如 x | y)，您可以使用以下类型:

If the signature is always the same as in your post, whereas the value of type must match the associating property (e.g. x | y), you can get away with the following type:

type TypeToValue<T extends Type> = { [P in T]: number } & { type: T }

可以如下使用:

declare function getRecord<T extends Type>(type: T, id: string): TypeToValue<T>

const { type, x } = getRecord('x', 'aa');

无法从与 type 参数匹配的联合类型中获取相应的类型.TS Playground

There is no way to get the corresponding type from the union type that matches the type argument. TS Playground

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