如何在 TypeScript 或 Flow 中删除未定义的继承类型的联合类型细化

Question

我继承了一个类型，正在从库中导入，我想知道如何才能删除它对空值或未定义值的允许。

type Foo = {
    baz: string
}

// Bar type is inherited and I would like to kill the union with undefined so I expect foo to never be falsey.
type Bar = {
    foo?: Foo
}
const bar: Bar = {foo: {baz: 'baz'}};
// this destructuring issues an error because it allows for the possibility of it being undefined and undefined can't be destructured. And I can't conditionally exit since I'm using React hooks and I'd be violating the hooks should not be used conditionally rule
const {baz} = bar.foo;

Answer 1

内置的打字稿中有一个实用程序类型叫做Required，你可以像下面这样使用它

type Foo = {
  baz: string
}

type Bar = {
  foo?: Foo
}

type StrictBar = Required<Bar>
// type StrictBar = { foo: Foo } inferred

如何在 TypeScript 或 Flow 中删除未定义的继承类型的联合类型细化

how to remove the union type refinement of of inherited type with undefined in TypeScript or Flow

typescript

flowtype