Reactjs Dynamic redux中间件typescript错误：预期类型来自属性'；类型'；在类型'；行动'；

我正在使用Typescript和react-redux，并尝试为我的API请求制作定制的动态中间件

这是我的密码：

import { Dispatch } from "react";
import { errorHandler } from "../components/Layout/SnackBar/alert";
import { API } from "../_helpers/api";
import { getTheTime } from "../_helpers/constants";
import { AppActions, AppState } from "../_types";

export const BankPages = {
  posts: { name: "POSTS", api: "/post/v2" },
  lessons: { name: "LESSONS", api: "/lesson/v2" },
  guides: { name: "GUIDES", api: "/lesson/v2/guide" },
  courses: { name: "COURSES", api: "/lesson/v2/course" },
  exercises: { name: "EXERCISES", api: "/lesson/v2?course=125" },
};

export const requestBank = (page: keyof typeof BankPages) => (
  dispatch: Dispatch<AppActions>,
  getState: () => AppState
) => {
  const bankState = getState().bank[page];
  const currentTime = getTheTime();
  const resetTime = currentTime - bankState.nextLoad;
  if (bankState.data && resetTime <= 0) {
    return;
  }
  dispatch({ type: `REQUEST_${BankPages[page].name}` });

  API.get(BankPages[page].api)
    .then((res) =>
      dispatch({
        type: `SUCCESS_${BankPages[page].name}`,
        payload: {
          data: res.data,
          nextLoad: getTheTime(10),
        },
      })
    )
    .catch((err) => errorHandler(err, `FAILURE_${BankPages[page].name}`));
};

一切都很完美，我知道这是因为我声明了

typeof REQUEST\u POSTS

---

如何修复此错误？

经过一些搜索，我找到了一个用于在Typescript 4.1+中使用以下函数生成动态字符串的方法：

function makeType<NS extends string, N extends string>(namespace: NS, name: N) {
  return namespace + name as `${NS}${N}`
}
// should be used like this
const sampleType = makeType('REQUEST_', 'POSTS');
// return "REQUEST_POSTS"

export const BankPages = {
  posts: { name: "POSTS", api: "/post/v2" } ,
  lessons: { name: "LESSONS", api: "/lesson/v2" } ,
  guides: { name: "GUIDES", api: "/lesson/v2/guide" } ,
  courses: { name: "COURSES", api: "/lesson/v2/course" } ,
  exercises: { name: "EXERCISES", api: "/lesson/v2?course=125" },
} as const;

dispatch({ type: makeType("REQUEST_", BankPages[page].name) });

当你键入

BankPages.post.name

时，使用此技巧可以显示

“POSTS”

而不是

string

动态调度应该是这样的：

function makeType<NS extends string, N extends string>(namespace: NS, name: N) {
  return namespace + name as `${NS}${N}`
}
// should be used like this
const sampleType = makeType('REQUEST_', 'POSTS');
// return "REQUEST_POSTS"

export const BankPages = {
  posts: { name: "POSTS", api: "/post/v2" } ,
  lessons: { name: "LESSONS", api: "/lesson/v2" } ,
  guides: { name: "GUIDES", api: "/lesson/v2/guide" } ,
  courses: { name: "COURSES", api: "/lesson/v2/course" } ,
  exercises: { name: "EXERCISES", api: "/lesson/v2?course=125" },
} as const;

dispatch({ type: makeType("REQUEST_", BankPages[page].name) });

---

我觉得

typeof REQUEST\u POSTS

的界面很好。我注意到的是你如何“构建”你的类型以供分派<代码>键入：“SUCCESS_${BankPages[page]}”这在我看来是错误的，它可能不是一个有用的字符串。@伊恩，谢谢你的评论，你喜欢用哪种方式制作动态字符串？重点不是偏好

BankPages[page]

不是字符串，而是对象。我不知道您是如何定义AppActions的，但它可能不需要字符串化对象。您可能想添加

。name

谢谢您的帮助

您可能想添加。name

我会处理它。