为什么我不能使这种类型通用?输入“x"不能用于索引类型“y"；ts(2536)

问题描述

我正在尝试编写一个泛型类型，它采用根级属性名称并返回嵌套在其下的属性的联合类型.例如:

I'm trying to write a generic type which takes a root-level property name and returns a union type of a property nested underneath it. For example:

interface operations {
  updateSomething: {
    "201": {
      schema: number;
    };
    "400": {
      schema: string;
    };
  };
}

如果我想获得模式"对于 updateSomething 类型，它应该解析为 number |字符串.非通用版本工作正常:

If I want to get the "schemas" for the type updateSomething, it should resolve to number | string. The non-generic version works fine:

type UpdateSomethingSchema =
  operations["updateSomething"][keyof operations["updateSomething"]]["schema"];

// string | number ✓

我尝试编写泛型类型是:

My attempt at writing a generic type is:

type SchemaOf<
  O extends keyof operations
> = operations[O][keyof operations[O]]["schema"];

但这给了我一个错误:

Type '"schema"' cannot be used to index type 'operations[O][keyof operations[O]]'.ts(2536)

有趣的是，如果我忽略错误，该类型似乎确实有效:

Interestingly, if I ignore the error, the type does seem to work:

type UpdateSomethingSchema = SchemaOf<"updateSomething">;

// string | number ✓

我做错了什么，还是 TypeScript 的限制?

Am I doing something wrong, or is this a limitation of TypeScript?

推荐答案

你可以借助分布式条件类型来实现:

You can achieve it with help of distributive conditional types:

type Schema<T> = {
  schema: T
}


interface operations {
  updateSomething: {
    "201": Schema<number>;
    "400": Schema<string>;
  };
}

type SchemaOf<
  O extends keyof operations
  > = operations[O][keyof operations[O]] extends Schema<infer S> ? S : never

type Result = SchemaOf<'updateSomething'> // string | number

IF operations[O][keyof operations[O]] 推断具有 schema 属性的对象，TypeScript 能够推断出 schema 的类型:T 并且由于分布性，它将返回联合类型.

IF operations[O][keyof operations[O]] infers to object with schema property, TypeScript is able to infer the type of schema: T and because of distributivity, it will return a union type.

分布式条件类型文档

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