Typescript 是否支持互斥类型? [英] Does Typescript support mutually exclusive types?

问题描述

我有一个接受参数的方法.我希望 Typescript 验证传入的对象(在 typescript 编译时，我理解运行时是一种不同的动物)仅满足一个允许的接口.

I have a method that takes a parameter. I would like Typescript to verify that the object being passed in (at typescript compile-time, I understand run-time is a different animal) only satisfies one of the allowed interfaces.

示例:

interface Person {ethnicity: string;}
interface Pet {breed: string;}
function getOrigin(value: Person ^ Pet){...}

getOrigin({}); //Error
getOrigin({ethnicity: 'abc'}); //OK
getOrigin({breed: 'def'}); //OK
getOrigin({ethnicity: 'abc', breed: 'def'});//Error

我意识到 Person ^ Pet 不是有效的 Typescript，但这是我想尝试的第一件事，而且看起来很合理.

I realize that Person ^ Pet is not valid Typescript, but it's the first thing I thought to try and seemed reasonable.

推荐答案

如 这个问题，你可以使用条件类型 写一个 XOR 类型:

As proposed in this issue, you could use conditional types to write a XOR type:

type Without<T, U> = { [P in Exclude<keyof T, keyof U>]?: never };
type XOR<T, U> = (T | U) extends object ? (Without<T, U> & U) | (Without<U, T> & T) : T | U;

现在您的示例有效:

interface Person {ethnicity: string;}
interface Pet {breed: string;}
function getOrigin(value: XOR<Person, Pet>) { /* ... */}

getOrigin({}); //Error
getOrigin({ethnicity: 'abc'}); //OK
getOrigin({breed: 'def'}); //OK
getOrigin({ethnicity: 'abc', breed: 'def'});//Error

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