如何创建需要设置单个属性的类部分

Question

我们的结构如下所示：

export type LinkRestSource = {
    model: string;
    rel?: string;
    title?: string;
} | {
    model?: string;
    rel: string;
    title?: string;
} | {
    model?: string;
    rel?: string;
    title: string;
};

这几乎等同于说

type LinkRestSource = Partial<{model: string, rel: string, title: string}>

只是这将允许传入一个空对象，而初始类型需要传入一个属性

如何创建像 Part 这样的泛型，但它的行为类似于上面的结构？

Answer 1

jcalz解决方案的更简单版本：

类型 至少一个

所以整个实现变得

js prettyprint-override">type FullLinkRestSource = {
  model: string;
  rel: string;
  title: string;
}

type AtLeastOne<T> = { [K in keyof T]: Pick<T, K> }[keyof T]
type LinkRestSource = AtLeastOne<FullLinkRestSource>

const okay0: LinkRestSource = { model: 'a', rel: 'b', title: 'c' }
const okay1: LinkRestSource = { model: 'a', rel: 'b' }
const okay2: LinkRestSource = { model: 'a' }
const okay3: LinkRestSource = { rel: 'b' }
const okay4: LinkRestSource = { title: 'c' }

const error0: LinkRestSource = {} // missing property
const error1: LinkRestSource = { model: 'a', title: 'c' } // excess property on string literal

这是TS游乐场的链接来试试

Answer 2

如果你知道你想要哪些属性，还有另一个解决方案。

type AtLeast<T, K extends keyof T> = Partial<T> & Pick<T, K>

这也将允许您锁定一个类型的多个密钥，例如。

type LinkRestSource = AtLeast<T, 'model' | 'rel'>

Answer 3

我想我有一个解决方案。您正在寻找一个采用T类型并生成至少包含T中一个属性的相关类型的东西。也就是说，它就像＜code＞Partial

如果是，这里是：

type AtLeastOne<T, U = {[K in keyof T]: Pick<T, K> }> = Partial<T> & U[keyof U]

要分析它：首先，＜code＞AtLeastOne

嗯，这可能会让人困惑。让我们逐步了解它如何在以下具体类型上运行：

type FullLinkRestSource = {
  model: string;
  rel: string;
  title: string;
}

type LinkRestSource = AtLeastOne<FullLinkRestSource>

那扩展到

type LinkRestSource = AtLeastOne<FullLinkRestSource, {
  [K in keyof FullLinkRestSource]: Pick<FullLinkRestSource, K>
}>

或

type LinkRestSource = AtLeastOne<FullLinkRestSource, {
  model: Pick<FullLinkRestSource, 'model'>,
  rel: Pick<FullLinkRestSource, 'rel'>,
  title: Pick<FullLinkRestSource, 'title'>
}>

或

type LinkRestSource = AtLeastOne<FullLinkRestSource, {
  model: {model: string},
  rel: {rel: string},
  title: {title: string}>
}>

或

type LinkRestSource = Partial<FullLinkRestSource> & {
  model: {model: string},
  rel: {rel: string},
  title: {title: string}>
}[keyof {
  model: {model: string},
  rel: {rel: string},
  title: {title: string}>
}]

或

type LinkRestSource = Partial<FullLinkRestSource> & {
  model: {model: string},
  rel: {rel: string},
  title: {title: string}>
}['model' | 'rel' | 'title']

或

type LinkRestSource = Partial<FullLinkRestSource> &
  ({model: string} | {rel: string} | {title: string})

或

type LinkRestSource = {model?: string, rel?: string, title?: string} & 
  ({model: string} | {rel: string} | {title: string})

或

type LinkRestSource = { model: string, rel?: string, title?: string } 
  | {model?: string, rel: string, title?: string} 
  | {model?: string, rel?: string, title: string}

我想，这就是你想要的。

您可以测试它：

const okay0: LinkRestSource = { model: 'a', rel: 'b', title: 'c' }
const okay1: LinkRestSource = { model: 'a', rel: 'b' }
const okay2: LinkRestSource = { model: 'a' }
const okay3: LinkRestSource = { rel: 'b' }
const okay4: LinkRestSource = { title: 'c' }

const error0: LinkRestSource = {} // missing property
const error1: LinkRestSource = { model: 'a', titel: 'c' } // excess property on string literal

那么，这对你有用吗？祝你好运！