typescript - 如何在打字稿中不使用选项（？）删除属性

Question

我有关于删除属性的问题。例如：

type Person<GN>  = {
    getName: GN extends never ? never :  GN,
}

const foo =  <GN>(person: Person<GN>) => person

const first = foo({}) // should work

const second = foo({
    getName: (name: string) => name,
})

在这种情况下，首先不需要 getName 属性。我该如何解决。

---

使用可选的“？” 属性会导致输出不清晰。例如

type Person<GN>  = {
    getName?: GN,
}

const foo =  <GN>(person: Person<GN>) => person

const first = foo({}) // should work
first.getName // shouldn't exist 'getName' => but still have

const second = foo({
    getName: (name: string) => name,
})

second.getName // should exist getName but in this case is optional property

我怎样才能清楚地输出？谢谢阅读。

标签： typescript

Answer 1

因此，您希望firsttype 被推断为{}，并且secondtype 为{ getName: (name: string) => string; }?

这是一种方法：

type Person<GN> = { getName: GN }

const foo = <GN, P extends Partial<Person<GN>>>(person: P) => person;

const first = foo({}) // const first: {}

const second = foo({  // const second: { getName: (name: string) => string; }
    getName: (name: string) => name,
})