typescript - 定义另一个属性时需要一个属性(有区别的联合)

Question

我想编写一种类型，它允许使用一些始终需要的核心属性和一个可选属性来创建对象 imageUrl定义为字符串时表示 imageAltText也需要作为字符串，

{
 id: "test",
 imageUrl: "test",
 imageAltText: "test"
}

我也希望它能够工作，以便当 imageUrl没有定义我不指望 imageAltText被定义为。

{
 id: "test"
}

我已将类型定义如下，

(
  {
    /** image url for the banner of the card. Will display a blue background without an image */
    imageUrl?: undefined;
  } 
| 
  {
    /** image url for the banner of the card. Will display a blue background without an image */
    imageUrl: string;
      /** alternative text for the banner image */
    imageAltText: string;
  }
)

However, by making `imageUrl` optional `?` typescript allows me to write `imageAltText` even when `imageUrl` is undefined. 

Answer 1

在这里你有:

type RequiredParams<T = {}> = {
    id: string
} & T

type UnionParams =
    | RequiredParams<{ imageUrl: string; imageAltText: string; }>
    | RequiredParams

type UnionKeys<T> = T extends T ? keyof T : never;
type StrictUnionHelper<T, TAll> =
    T extends any
    ? T & Partial<Record<Exclude<UnionKeys<TAll>, keyof T>, never>> : never;

type StrictUnion<T> = StrictUnionHelper<T, T>

type Result = StrictUnion<UnionParams>

const a: Result = { id: 'sdf' }; // ok
const b: Result = { id: 'sdf', imageUrl: 'sdf', imageAltText: 'sdf' } // ok
const c: Result = { id: 'sdf', imageUrl: 'sdf' } // error
const d: Result = { id: 'sdf', imageAltText: 'sdf' } // error

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