英文:
How to get union of value types of a nested object?
问题
以下是代码的翻译部分：
给定这样的对象：```tsconst 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;
<details><summary>英文:</summary>Given an object like this:```tsconst deepNestedObject = {a: "value",b: {c: "nestedvalue"d: {e: "deepNestedValue"}}}
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;
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;
英文:
Extract all the strings from the values:
type EdgeValueUnion = Extract<DeepValueOf<typeof deepNestedObject>, string>;
You also need a const assertion to make sure the strings are inferred as literals instead of the general string type:
const deepNestedObject = {a: "value",b: {c: "nestedvalue",d: {e: "deepNestedValue",},},} as const;
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