重嵌套对象的TypeScript类型

Question

我得到了一棵树，我正试图以递归方式呈现

树变量只是一个例子，它可能会变得更大，这取决于应用程序获得的数据。

我如何才能让TypeScript对这棵树上的类型感到满意，即使我不知道嵌套将如何得到？

const tree = {
  people: ['Managing Director'],
  children: {
    people: ['Operations Director', 'Head of Client Services'],
    children: {
      people: ['Senior Developer']
    }
  }
}

interface IList {
  people: string[],
  children: string[]
}

interface IData {
  data: IList[]
}

const List: FC<IData> = ({ data }) => (
  <ul>
    {data.people.map((person) => ( <li>{person}</li> ))}
    {!data.children ? null : <List data={data.children} />}
  </ul>
)

function App() {
  return (
    <>
      <List data={tree} />
    </>
  )
}

当我在codesandbox上做这件事时，它是有效的，但有警告，如果我在我的配置上做，我会得到

`Property 'people' does not exist on type 'IList[]'`

EXAMPLE

Answer 1

您需要将children属性设置为可选和递归类型：

type Tree = {
    people: Array<string>;
    children?: Tree;
}

const tree: Tree = {
  people: ['Managing Director'],
  children: {
    people: ['Operations Director', 'Head of Client Services'],
    children: {
      people: ['Senior Developer']
    }
  }
}

然后，List可以接受Tree类型的属性并递归呈现它。

const List = ({ data }: { data: Tree }) => (
    <ul>
        {data.people.map((person) => (<li>{person}</li>))}
        {!data.children ? null : <List data={data.children} />}
    </ul>
)