在通用声明中破坏的用途是什么？

发布于 2025-01-24 18:14:35 字数 364 浏览 1 评论 0原文

破坏可能是错误的术语...

在我正在工作的代码库中找到了这一点。这令人困惑。

await apolloClient
    .query<{ Workorder: Workorder }>({
      query: WORKORDER_QUERY,
      variables: { uuid: id },
    })

我看不到{workorder：workorder}的点。这里发生了什么？该vs .query＆lt; workorder＆gt;的值是什么？

Workorder是一个大约30个属性的接口。

原文

Destructuring might be the wrong term...

Found this in the codebase I'm working in. It's perplexing.

await apolloClient
    .query<{ Workorder: Workorder }>({
      query: WORKORDER_QUERY,
      variables: { uuid: id },
    })

I don't see the point of { Workorder: Workorder }. What is happening here? What is the value of that vs .query<Workorder>?

Workorder is an interface with about 30 properties.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

雾里花 2025-01-31 18:14:35

这不是破坏性的。这只是意味着要传递的类型参数是具有工作表属性的对象，并且该属性的值是工作派。

例如，给定的：

type Workorder = { order: boolean };
declare function fn<T>(): T;

const result = fn<Workorder>();

结果是类型：而

{
  order: boolean
}

如果执行结果，则类型为类型，而

const result = fn<{ Workorder: Workorder }>();

结果现在是类型：

{
  Workorder: {
    order: boolean
  }
}

不是.query 做同样的事情，但这会让您了解区别是什么。执行＆lt; {workorder：workorder}＆gt;只需将工作端包装在对象中，然后将其作为类型参数传递。

It's not destructuring. It just means that the type argument being passed is an object with a Workorder property, and that the value for that property is a Workorder.

For example, given:

type Workorder = { order: boolean };
declare function fn<T>(): T;

const result = fn<Workorder>();

where the type parameter is the same type that's returned, the result is of type:

{
  order: boolean
}

Whereas if you do

const result = fn<{ Workorder: Workorder }>();

the result is now of the type:

{
  Workorder: {
    order: boolean
  }
}

Not that .query does the same thing, but that'll give you an idea of what the difference is. Doing <{ Workorder: Workorder }> just wraps the Workorder in an object and passes it as a type parameter.

回复收藏 0 原文

~没有更多了~