[英]How to get a literal type from a class property (after passing the class as argument) for a computed property name?
a moment ago I asked this question and now I have a follow up one:)刚才我问了这个问题,现在我有一个后续问题:)
Please consider the following code: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 recognizes data as the following type: TS 将数据识别为以下类型:
const data: {} | undefined
I'd like TS to know that data has a property books我想让 TS 知道 data 有一个属性 books
const data: {
books: Book[];
} | undefined
Is it possible?可能吗?
This is possible with a combination of index access types and mapped types ( playground ):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
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.