问题

以下是代码的翻译部分：

给定这样的对象：
```ts
const deepNestedObject = {
  a: "value",
  b: {
    c: "nestedvalue",
    d: {
       e: "deepNestedValue"
    }
  }
}

如何生成一个联合类型，仅限于边缘的非对象值

type EdgeValueUnion = DeepValueOf<typeof deepNestedObject>; // 应该等同于 "value" | "nestedValue" | "deepNestedValue"

到目前为止，我尝试过这个：

type ValueOf<T> = T[keyof T];

type DeepValueOf<
  T extends Record<string, unknown>,
  Key = keyof T
> = Key extends string
  ? T[Key] extends Record<string, unknown>
    ? DeepValueOf<T[Key]>
    : ValueOf<T>
  : never;

但它还不太对，因为它仍然允许树中的对象值通过。例如：

const edgeValue: EdgeValueUnion = deepNestedObject.b;

示例TS沙箱

<details>
<summary>英文:</summary>

Given an object like this:
```ts
const deepNestedObject = {
  a: &quot;value&quot;,
  b: {
    c: &quot;nestedvalue&quot;
    d: {
       e: &quot;deepNestedValue&quot;
    }
  }
}

How do I generate a union type of specifically only the non-object values at the edge

type EdgeValueUnion = DeepValueOf<typeof deepNestedObject>; // Should equate to "value" | "nestedValue" | "deepNestedValue"

So far I've tried this:

type ValueOf<T> = T[keyof T];

type DeepValueOf<
  T extends Record<string, unknown>,
  Key = keyof T
> = Key extends string
  ? T[Key] extends Record<string, unknown>
    ? DeepValueOf<T[Key]>
    : ValueOf<T>
  : never;

but it's not quite right because it still allows object values in the tree to pass. e.g.

const edgeValue: EdgeValueUnion = deepNestedObject.b;

Sample TS sandbox

答案1

得分: 1

从值中提取所有字符串：

type EdgeValueUnion = Extract<DeepValueOf<typeof deepNestedObject>, string>;

还需要使用 const 断言来确保这些字符串被推断为字面量，而不是通用的 string 类型：

const deepNestedObject = {
    a: "value",
    b: {
        c: "nestedvalue",
        d: {
            e: "deepNestedValue",
        },
    },
} as const;

Playground

type EdgeValueUnion = Extract<DeepValueOf<typeof deepNestedObject>, string>;

const deepNestedObject = {
    a: "value",
    b: {
        c: "nestedvalue",
        d: {
            e: "deepNestedValue",
        },
    },
} as const;