问题

我有一个类似这样的想法：

```ts
type Foo<T, K extends string> = K extends "isDirty"
  ? never
  : {
      [P in K]: T;
      isDirty: boolean;
    };

但是 TypeScript 仍然不知道 K 永远不会是 "isDirty"。

我想做的是，允许任何键值，所以它可以是 foo、bar、potatoes，并且将其类型设为 T。但同时也有属性 isDirty: boolean，这样 K 可以拥有任何值，但不包括这个。

一些示例：

const foo: Foo<number, "bar"> = {
  bar: 5,
  isDirty: false
}

const potatoes: Foo<number[], "potatoes"> = {
  potatoes: [1,6,15],
  isDirty: false
}

在 TypeScript 中是否可能实现这样的功能？

<details>
<summary>英文:</summary>

I had in mind something like this:

```ts
type Foo&lt;T, K extends string&gt; = K extends &quot;isDirty&quot;
  ? never
  : {
      [P in K]: T;
      isDirty: boolean;
    };

But Typescript still doesn't know that K will never be `"isDirty".

What I wanted to do is, allow any key value, so it could be foo, bar, potatoes and have that be of type T. But also have the property isDirty: boolean, so K can have any value but that.

Some examples:

const foo: Foo<number, "bar"> = {
  bar: 5,
  isDirty: false
}

const potatoes: Foo<number[], "potatoes"> = {
  potatoes: [1,6,15],
  isDirty: false
}

Is this possible to do in Typescript?

答案1

得分: 1

You can use an intersection of {[P in K]: T} and {isDirty: boolean}:

type Foo<T, K extends string> = K extends "isDirty"
    ? never
    : {
          [P in K]: T;
      } & {
          isDirty: boolean;
      };

const foo: Foo<number, "bar"> = {
    bar: 5,
    isDirty: false,
};

const potatoes: Foo<number[], "potatoes"> = {
    potatoes: [1, 6, 15],
    isDirty: false,
};

Playground link

type Foo<T, K extends string> = K extends "isDirty"
    ? never
    : {
          [P in K]: T;
      } & {
          isDirty: boolean;
      };

const foo: Foo<number, "bar"> = {
    bar: 5,
    isDirty: false,
};

const potatoes: Foo<number[], "potatoes"> = {
    potatoes: [1, 6, 15],
    isDirty: false,
};