typescript - 如何避免TypeScript中的动态keyof对象分配错误

Question

假设我们的 TypeScript 代码如下所示:

type User = {
  id: number,
  name: string,
}

let user1: User = {id: 123, name: "Hello"};
let user2: User = {id: 456, name: "World"};

let keys: (keyof User)[] = ["id", "name"];

for (let key of keys) {
  user1[key] = user2[key];
}

这给出了错误

Type 'string | number' is not assignable to type 'never'.

对于声明

user1[key] = user2[key];

如果我们将 keys 的定义改为

let keys: string[] = ["id", "name"];

错误消失了，但我们失去了类型安全。

有什么方法可以在保持类型安全的同时避免这个错误？

Answer 1

这里没有避免类型断言的好方法。在 TS 的最新版本中(我认为是 3.5 后)，当通过索引写入时，写入的值必须与键指定的所有可能的属性值兼容。在你的情况下，这将是 number & string ，它减少到 never 因此错误。

根本原因是 TS 不只跟踪类型的变量，因此就类型而言，您的示例与:

let key1 = 'id' as  keyof User;
let key2 = 'name' as  keyof User;
//Obvious error
user1[key1] = user2[key2] // same error, TS can't distingusih between this and your user1[key] = user2[key]

最简单的解决方案是使用类型断言，如果在您的情况下您确定这没问题:

type User = {
  id: number,
  name: string,
}


let user1: User = { id: 123, name: "Hello" };
let user2: User = { id: 456, name: "World" };
for (let key of keys) {
  user1[key] = user2[key] as never
}

Play

或者(但不再是类型安全的)您可以使用一个小漏洞，其中 T[K] 可分配给索引值:

type User = {
  id: number,
  name: string,
}


let user1: User = { id: 123, name: "Hello" };
let user2: User = { id: 456, name: "World" };

let keys: (keyof User)[] = ["id", "name"];

for (let key of keys) {
  set(user1, key, user2[key])
}

Play