Typescript typescribe不能使用值来索引对象类型，该对象类型满足具有该值的另一个类型'

Paths<T>类型的问题在于T是generic类型constrained到typeof apiPaths.这意味着它可能是typeof apiPaths中的一些extension，而typeof apiPaths中不存在额外属性，并且没有任何要求这些属性具有path属性. 例如:

type ApiPathsWithRaisins = typeof apiPaths & {raisins: true};
type Oops = Paths<ApiPathsWithRaisins>; 

Paths<T>定义为

type Paths<T extends typeof apiPaths = typeof apiPaths> = {
  [K in keyof T]: T[K] extends { chain: Chain }
  ? `${T[K]['path']}${T[K]['chain']}/`
  : `${T[K]['path']}`
}[keyof T]

那么对于Paths<ApiPathsWithRaisins>，K将超过"getToken" | "getRank" | "raisins". 当K是raisins，T[K]是true，而true没有一个名为path的已知成员.因此，你不能肯定地假设T[K]将具有path属性.

这就是错误的来源.同样，已知typeof apiPaths在每个键上都有path属性，但T没有.

你可以用任何方法来解决它. 如果你只关心计算Paths<typeof apiPaths>，那么实际上根本不需要泛型.只要用typeof apiPaths代替T就可以了:

type MyApiPaths = typeof apiPaths;
type Paths = {
  [K in keyof MyApiPaths]: MyApiPaths[K] extends { chain: Chain }
  ? `${MyApiPaths[K]['path']}${MyApiPaths[K]['chain']}/`
  : `${MyApiPaths[K]['path']}`
}[keyof MyApiPaths]
// type Paths = "rank/" | "tokens/eth/"

或者甚至像

type Paths = keyof { [T in MyApiPaths[keyof MyApiPaths]
  as T extends { chain: infer C extends Chain } ? `${T["path"]}${C}/` : T['path']]: 0
}
// type Paths = "rank/" | "tokens/eth/"

或者，如果你确实想使用泛型，你只需要确保它被适当地约束到每个属性都有一个类似string的属性:

type GoodPaths<T extends Record<keyof T, { path: string }> = typeof apiPaths> = {
  [K in keyof T]: T[K] extends { chain: Chain }
  ? `${T[K]['path']}${T[K]['chain']}/`
  : `${T[K]['path']}`
}[keyof T]
type OK = GoodPaths;
// type OK = "rank/" | "tokens/eth/"

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