在Typescript中具有特定类型的任何属性的任何属性的泛型类型

Question

大家好。:)

作为这个问题的后续，我们有以下代码:

interface SubThing {
name: string;
}
interface Thing {
subThing: SubThing; // or any other property of type SubThing like in UpperThing
}
interface UpperThing {
thing: Thing
}
interface ThingMapA {
map: { thing: { subThing: { name : 'subThing name' } } };
}
interface ThingMapB {
map: { thing: { otherThing: { name : 'subThing name' } } };
}
class Handler<T extends Record<keyof T, Record<keyof T, Thing>>> {}
const handler = new Handler<ThingMapA>(); // error
const handler = new Handler<ThingMapB>(); // error

我喜欢Handler接受任何具有UpperThing类型属性(理想情况下至少有一个)的类，并且具有Thing类型的任何属性(理想情况下至少有一个)。但ThingMap不被识别。这会导致错误。有什么建议吗?😊

---

这是另一个基于@jcalz

方法的例子

interface SubThing {
name: string;
}
interface Thing {
subThing: SubThing;
}
interface UpperThing {
thing: Thing
}
interface ThingMapA {
map: { thing: { subThing: { name: 'subThing name' } } };
}
interface ThingMapB {
map: { otherThing: { subThing: { name: 'subThing name' } } };
}
interface ThingMapC {
map: { thing: { otherSubThing: { name: 'subThing name' } } };
}
interface ThingMapD {
map: { otherThing: { otherSubThing: { name: 'subThing name' } } };
}
class Handler<T extends { [K in keyof T]: { [P in keyof T[K]]: Thing } }> { }
const handlerA = new Handler<ThingMapA>(); // okay
const handlerB = new Handler<ThingMapB>(); // okay
const handlerC = new Handler<ThingMapC>(); // error
const handlerD = new Handler<ThingMapD>(); // error

Answer 1

如果您需要约束类型T，以便每个子属性都是类型SubThing，那么您将需要编写涉及嵌套映射类型的适当深度的递归约束:

class Handler<T extends {
[K in keyof T]: {
[P in keyof T[K]]: {
[Q in keyof T[K][P]]: SubThing
}
}
}> { }

让我们测试一下:

const handlerA = new Handler<ThingMapA>(); // okay
const handlerB = new Handler<ThingMapB>(); // okay
const handlerC = new Handler<ThingMapC>(); // okay
const handlerD = new Handler<ThingMapD>(); // okay
const badhandler = new Handler<{
a: { b: { c: { name: "" } } },
d: { e: { f: { name: 123 } } } // error!
}>
//   The types of 'd.e.f.name' are incompatible between these types.

看起来不错。

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