fp-ts/typescript 避免嵌套管道

Question

在打字稿中使用 fp-ts 时如何避免嵌套管道？做记号？这是我想避免的嵌套管道示例

  pipe(
  userId,
  O.fold(
    () => setUser('No user found'),
    (uId: string) => {
      fetchUser(uId).then((response: Either<NonEmptyArray<string>, User>) => {
        pipe(
          response,
          E.fold(
            () => setUser('Unable to fetch user'),
            (user: User) => setUser(JSON.stringify(user, null, 2))
          )
        );
      });
    }
  )
);

Answer 1

一种方法是尽可能延迟fold，以避免不必要的嵌套。

代数效应可以在不需要知道值是否存在或操作是否失败的情况下组合。

例如，如果未找到用户，则使用 chain 转换的 TaskEither 将保留第一个错误。否则，它包含fetchUser 错误或成功案例中的User 数据。

Working example

import { pipeable as P, option as O, taskEither as TE, nonEmptyArray as NA } from "fp-ts";

type User = { name: string };

// global state (side effect)
let user: string | undefined = undefined;
const setUser = (usr: string) => (user = usr);

// given input
const userId: O.Option<string> = O.some("Arun");

const fetchUser: (uid: string) => TE.TaskEither<NA.NonEmptyArray<string>, User> = uid =>
  TE.taskEither.of({ name: "Arun" });
  // An error case would be: TE.left(NA.of("Unable to fetch user"))  

const res = P.pipe(
  userId,
  TE.fromOption(() => NA.of("No user found")),
  TE.chain(fetchUser),
  TE.map(user => JSON.stringify(user, null, 2)),
  TE.fold( // we fold at the very end before applying the effect
    err => TE.taskEither.fromIO(() => { setUser(err[0]); }),
    user => TE.taskEither.fromIO(() => { setUser(JSON.stringify(user, null, 2)); })
  ),
  TE.chain(() => TE.taskEither.fromIO(() => { console.log(user); }))
);

// run the effect
res();

PS：我在这里假设您的fetchUser 是一个异步操作，它创建TaskEither。如果需要，您可以将其切换回 Either。