TypeScript 如何实现属性的类型取决于另外一个属性的值？

This topic created in 1576 days ago, the information mentioned may be changed or developed.

代码如下，想要实现 Message 中：playload 的类型取决于 type 的值

下面已有的写法是不行的。

 interface PayloadMap { join: PeerInfo; offer: RTCSessionDescriptionInit; answer: RTCSessionDescriptionInit; icecandidate: RTCIceCandidateInit; leave: PeerInfo; } // 客户端发送，服务端接受的数据格式 interface Message { type: 'join' | 'offer' | 'answer' | 'icecandidate' | 'leave'; nick: string; id: string; // playload 的类型取决于 type 的值 payload: PayloadMap[Message['type']]; }

比如：

const m: Message = { type: 'join', nick: 'Fuck', id: '123456', payload: { // 此时 payload 应该是 PeerInfo 类型 } }

3 replies 2022-04-14 11:01:49 +08:00

rabbbit

Jan 21, 2022

interface Foo {

}

interface Bar {

}

interface PayloadMap {
　 join: Foo;
　 offer: Bar;
}

type Message = { [k in keyof PayloadMap]: {
　 type: k;
　 payload: PayloadMap[k];
} }[keyof PayloadMap];

const m: Message = {
　 type: 'join',
　 payload: {
　}
}

const n: Message = {
　 type: 'offer',
　 payload: {
　}
}

iPhone12

Jan 21, 2022

@rabbbit 学到了，感谢！

map 出一个 key 对应一个类型，然后联合。妙！

fifa666

Apr 14, 2022

interface PayloadMap {
join: 'join';
offer: RTCSessionDescriptionInit;
answer: RTCSessionDescriptionInit;
icecandidate: RTCIceCandidateInit;
leave: 'leave';
}

interface Message<T, P extends keyof T> {
type: P;
nick: string;
id: string;
payload: T[P];
}

const m: Message<PayloadMap, 'join'> = {
type: 'join',
nick: 'Fuck',
id: '123456',
payload: "join"
};