在严格的null检查中，可选属性内的属性的typescript类型

Question

Consider the following interface: 考虑以下接口：

interface MyInterface {
  parent?: {
    child: { ... }
  }
}

Now I want to access the type of 'child'. 现在，我要访问“孩子”的类型。 Normally I would do that: 通常我会这样做：

type ChildType = MyInterface['parent']['child']

But, when enabling strict-null-checks mode, I unfortunately get the following error: 但是，当启用strict-null-checks模式时，不幸的是出现以下错误：

TS2339: Property 'child' does not exist on type '{ child: { ... }} | undefined`.

That makes sense, because it's indeed not a property of undefined. 这是有道理的，因为它确实不是undefined的属性。

I tried using the non-null-assertion-operator: 我尝试使用non-null-assertion-operator：

type ChildType = MyInterface['parent']!['child']

But I got that error: 但是我得到了这个错误：

TS8020: JSDoc types can only be used inside documentation comments.

God knows what that means... 上帝知道那意味着什么...

Any way to get the type of 'child' here? 有什么办法在这里得到“孩子”的类型吗？

Answer 1

You just need to exclude undefined from the type of parent : 您只需要从parent类型中排除undefined ：

interface MyInterface {
  parent?: {
    child: { a: number }
  }
}

type ChildType = Exclude<MyInterface['parent'], undefined>['child']