Question

了解rest elements in tuple types并尝试弄清楚如何提取类型的通用部分：

type Attribute<Type> = { id: string, type?: Type };

type Position = { x: number, y: number };
let Position: Attribute<Position> = { id: "position" };

type Status = "active" | "inactive";
let Status: Attribute<Status> = { id: "status" };

我确定有一种方法可以编写条件类型，该条件类型会将各种Attribute<T>的元组映射到各种T的元组。

type AttributeTypes<Attributes extends Attribute<any>[]> =
   Attributes extends Attribute<infer T> ? T[] : never;

type Result = AttributeTypes<[typeof Position, typeof Status]> // should be `[Position, Status]`

但是我对推理步骤的理解不够充分，它总是以never分支结束。

最后一步将是编写一个将推断类型用作返回（Playground）一部分的函数：

function getAll<Attributes extends Attribute<any>[]>(
  ...attributes: Attributes
): AttributeTypes<Attributes> {
  return attributes.map(attribute => attribute.type);
}

let [position, status]: [Position, Status] = getAll(Position, Status);

Answer 1

条件类型没有理由在元组上工作，您的条件类型基本上可以解决问题[typeof Position, typeof Status] extends Attribute<infer T>，而问题AttributeTypes<typeof Position | typeof Status>显然没有，因此您永远都不会。

您可以将联合输入类型（Position[] | Status[]）中，然后将得到Attributes extends Array<Attribute<infer T>> ? T[] : never，它并不是您想要的（Play）

您还可以在条件类型（type Attribute<Type> = { id: string, type?: Type }; type Position = { x: number, y: number }; let Position: Attribute<Position> = { id: "position" }; type Status = "active" | "inactive"; let Status: Attribute<Status> = { id: "status" }; type AttributeTypes<Attributes extends Attribute<any>[]> = { [P in keyof Attributes]: Attributes[P] extends Attribute<infer T> ? T : never; } type Result = AttributeTypes<[typeof Position, typeof Status]> // is [Position, Status]）中使用数组，但不会在输入（Play）中保留元组结构

获得所需输出的最佳方法是使用映射类型。映射类型保留元组，同时允许您将元组的每种元素类型映射到结果元组中的新元素类型：

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Play

推断剩余元组类型的通用参数

1 个答案: