javascript - 如何使用 Typescript 通过类型而不是接口(interface)来建模对象？

Question

我有以下我无法控制的 JSON:

{
    "date": "2020-01-01T00:00:00",
    "a": 0,
    "b": 0,
    "c": 0
}

而不是指定接口(interface):

interface Foo {
    date: Date,
    "a": number,
    "b": number,
    "c": number
}

我希望能够像这样静态输入有效负载:

type Age = "a";
type Salary = "b";
type Height = "c"

type Foo = Record<"date", Date> | Record<Age | Salary | Height, number>;

尽管编译器看起来很高兴，但是当我尝试从对象获取日期时，我收到以下错误:

const foo = {
        "date": "2020-01-01T00:00:00",
        "a": 0,
        "b": 0,
        "c": 0
    } as Foo;

Element implicitly has 'any' type.

如果我删除 Record<Age | Salary | Height, number>然后错误就消失了。

Answer 1

您应该使用intersection而不是union :

type Foo = Record<"date", string> & Record<Age | Salary | Height, number>

Playground

<小时/>

** 我已将 date 类型更改为 string，因为您要为此字段分配字符串值