如何使typescript知道对象的单个属性的确切常量值

Question

我为此挣扎了一段时间。我的问题如下：我知道数据结构。

const array = [
{
foo: "foo1" as const,
bar: "any string"
baz: 2
},
{
foo: "foo2" as const,
bar: "any string 2"
baz: 2
}
] 

我可以简单地推断类型只有foo变量自动完成并具有精确值吗？

array[0].foo // has possible values of "foo1", "foo2"

但是

array[0].bar // has any string as a value

array[0].baz // has any number as a value

假设我有一些已知类型的接口：

interface Example {
name: ???? // how should i type it?
bar: string,
baz: number,
}

我试图在不为所有可能的名称声明类型的情况下对其进行存档。如果有任何建议，我将不胜感激。

---

编辑：

我尝试参数化示例界面：

interface Example<Name> { name: Name, ... rest of types }
type MapToNamed<T> = T extends { name: infer Name } ? Example <Name> : never
type TargetType = MapToNamed<typeof array[number]>;

但这迫使我使用as any as MapToNamed<typeof array>，我不确定这是否正确。

Answer 1

"foo1" as const称为const断言，用于只读值

对于您的情况，它不会影响您的类型声明

你可以像下面的一样定义Example

interface Example {
foo: string, //apply for all string types
bar: string,
baz: number,
}
const array: Example[] = [
{
foo: "foo1" as const, //consider as a read-only string
bar: "any string",
baz: 2
},
{
foo: "foo2" as const, //consider as a read-only string
bar: "any string 2",
baz: 2
}
] 
console.log({ array })

游乐场

如果您想要固定类型，如foo1和foo2。您可以使用枚举类型

enum Foo {
"foo1",
"foo2"
}
interface Example {
foo: keyof typeof Foo, //"foo1" | "foo2"
bar: string,
baz: number,
}
const array: Example[] = [
{
foo: "foo1" as const,
bar: "any string",
baz: 2
},
{
foo: "foo2" as const,
bar: "any string 2",
baz: 2
}
] 
console.log({ array })

游乐场