fp-ts/typescript避免嵌套管道

Question

在typescript中使用fp-ts时，如何避免嵌套管道？做符号吗？下面是我想要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

一种方法是尽可能地延迟fold，以避免不必要的嵌套。

代数效果可以在不需要知道值是否存在或操作失败的情况下合成。

例如，如果没有找到用户，使用chain转换的TaskEither将保留第一个错误。否则，它将包含成功案例中的fetchUser错误或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。