typescript - 如何从类属性（将类作为参数传递后）获取计算属性名称的文字类型？

Question

刚才我问了这个问题，现在我有一个跟进：）

请考虑以下代码：

import { ClassConstructor } from "class-transformer";
import { useQuery as useApolloQuery } from "@apollo/client";

class Book {
  readonly many = "books" as const;
  bookData: any;
}

export const useQueryWrapper = <T>(cls: ClassConstructor<T>, queryString) => {
  return useApolloQuery<{ [cls.prototype.many]: T[] }>(queryString);
};

const { data } = useQueryWrapper(Book, "..."); // Book or any other class with a literal `many` prop

TS 将数据识别为以下类型：

const data: {} | undefined

我想让 TS 知道 data 有一个属性 books

const data: {
    books: Book[];
} | undefined

可能吗？

Answer 1

这可以通过索引访问类型和映射类型（操场）的组合来实现：

class Book {
  readonly many = "books" as const;
  bookData: any;
}

class Page {
  readonly many = "pages" as const;
  bookData: any;
}

type ManyType = { readonly many: string };

type QueryResult<T extends ManyType> = {
  // using K in T["many"] to create an object with a key of value T["many"], e.g. "books", "pages", etc.
  [K in T["many"]]: T[]; 
};

type ClassConstructor<T> = new (...args: any[]) => T;

function useQueryWrapper<T extends ManyType>(
  cls: ClassConstructor<T>,
  queryString: string
): QueryResult<T> | undefined {
  return {} as any;
}

const books = useQueryWrapper(Book, "...")?.books; // Book[] | undefined
const pages = useQueryWrapper(Page, "...")?.pages; // Page[] | undefined