Question

我在删除属性方面遇到问题。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。我该怎么解决。

使用可选的“？”属性会导致输出不清楚。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

如何清楚地输出？感谢您的阅读。

Answer 1

因此，您希望将first类型推断为{}，并将second类型推断为{ getName: (name: string) => string; }吗？

这是一种实现方法：

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,
})

如何在打字稿中不使用option（？）的情况下删除属性

1 个答案: