fp-ts 如何处理不同权限类型的多个 Either

Question

当 Either 的右侧没有很好地对齐时，处理 Promise 和 Either 的最佳方法是什么？在这种情况下，我有三个非依赖的“先决条件”操作，表示为 Eithers（具有不同的右手类型）。如果他们都成功了，我不想进行第四次手术。如果三个都失败了，我不想再进行第四个手术了。

至此，我有一个正在编译的解决方案，但对可读性不满意。在这种情况下，肯定有更优雅的方法来处理多个 Either 吗？

    //promise of Either<ApiError, CustomerDTO>
    const customer = this.customerService.createCustomer(siteOrigin, createCustReq);

    //promise of Either<ApiError, LocationDTO>
    const location = this.locationService.getRetailOnlineLocation(siteOrigin);

    //promise of Either<ApiError, StationDTO>
    const station = this.stationService.getRetailOnlineStation(siteOrigin);
    
    //execute previous concurrently
    const locationAndStationAndCustomer = await Promise.all([location, station, customer]);

    const locationE = locationAndStationAndCustomer[0];
    const stationE = locationAndStationAndCustomer[1];
    const customerE = locationAndStationAndCustomer[2];


    //How to make this better?
    const stationAndLocationAndCustomer = E.fold(
      (apiErr: ApiError) => E.left(apiErr),
      (location: LocationDTO) => {
        return E.fold(
          (apiErr: ApiError) => E.left(apiErr),
          (station: StationDTO) =>
            E.right(
              E.fold(
                (err: ApiError) => E.left(err),
                (customer: CustomerDTO) =>
                  E.right({ location, station, customer })
              )(customerE)
            )
        )(stationE);
      }
    )(locationE);

Answer 1

我认为评论已经接近正确答案了。 sequenceT 是解决此类问题的正确方法。

import { sequenceT } from 'fp-ts/Apply'
import * as E from 'fp-ts/Either';

const seq = sequenceT(E.Apply);

return pipe(
  await Promise.all([location, station, customer]),
  seq, // [Either<...>, Either<...>, Either<...>] => Either<ApiError, [...]>
  // map is a bit less cumbersome. If the value was Left it returns Left
  // otherwise it calls the function which returns a new Right value from
  // what the function returns
  E.map(([loc, sta, cust]) => ({
    location: loc,
    station: sta,
    customer: cust,
  })), 
);

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