Typescript对于第三方API进行进一步封装的时候，如何保持类型一致？

Question

我需要使用ahooks（阿里巴巴的react hook库）的useRequest，然后根据我的需求，我想要封装一层通用处理，也就是包裹成一个自定义的hook。

ahooks的useRequest是有多个类型声明的，如下代码所示。

declare function useRequest<R = any, P extends any[] = any, U = any, UU extends U = any>(service: CombineService<R, P>, options: OptionsWithFormat<R, P, U, UU>): BaseResult<U, P>;

declare function useRequest<R = any, P extends any[] = any>(service: CombineService<R, P>, options?: BaseOptions<R, P>): BaseResult<R, P>;

declare function useRequest<R extends LoadMoreFormatReturn, RR>(service: CombineService<RR, LoadMoreParams<R>>, options: LoadMoreOptionsWithFormat<R, RR>): LoadMoreResult<R>;

declare function useRequest<R extends LoadMoreFormatReturn, RR extends R>(service: CombineService<R, LoadMoreParams<R>>, options: LoadMoreOptions<RR>): LoadMoreResult<R>;

declare function useRequest<R = any, Item = any, U extends Item = any>(service: CombineService<R, PaginatedParams>, options: PaginatedOptionsWithFormat<R, Item, U>): PaginatedResult<Item>;

declare function useRequest<R = any, Item = any, U extends Item = any>(service: CombineService<PaginatedFormatReturn<Item>, PaginatedParams>, options: BasePaginatedOptions<U>): PaginatedResult<Item>;

我自定义的hook，只是为了处理通用错误和通用请求的。。。其他传参和返回也应该和useRequest保持一致性。

我该如何给自定义的hook声明和useRequest一样的类型呢

Answer 1

我的就很简单，这样倒是可以套一层hook

import useRequest from "@umijs/use-request"
import { BaseOptions } from "@umijs/use-request/lib/types"

export function useNetworkTask<Result, Data, P extends any[] = any>(url: string, data: Data, options?: BaseOptions<Result, P>) {
    const result = useRequest<Result>(url, {
        ...options
    })
    return result
}