Redux-toolkit create action prepare 不添加 Typescript 的值

Question

我在此处遵循官方示例 redux-toolkit reference 并尝试按以下方式键入 PayloadAction :

import {
  createSlice,
  PayloadAction,
  nanoid,
} from '@reduxjs/toolkit'

type MyObjectType = {
  uuid: string
  anotherProp: string
  // ...
}

const slice = createSlice({
  name: 'oneSlice',
  initialState: {},
  reducers: {
    addSomething: {
      reducer(state, action: PayloadAction<{x: string; anotherProp: string}>) {
        const { x, uuid, anotherProp } = action.payload // got an error here on uuid
        // do something with { x, uuid, anotherProp }
        // I need to use it as index : state.something[uuid] = { uuid: uuid, anotherProp: anotherProps }
      },
      prepare(x: string, anotherProp: string) {
        return {
          payload: {
            uuid: nanoid(),
            x,
            anotherProp,
          },
        }
      },
    }
  }
})

但是我在尝试解构 action.payload 时遇到错误。如何推断 action.payload 的 uuid 属性？

我找到了一个奇怪的解决方案，我在 PayloadAction 中传递了整个对象类型并修改了 prepare()

中的 uuid

...

    addSomething: {
      reducer(state, action: PayloadAction<{x: string; myobject: MyObjectType}>) {
        const { x, myobject } = action.payload 
        // there is no problem here
      },
      prepare(x: string, myobject: MyObjectType) {
        myobject.uuid = nanoid()
        return {
          payload: {
            x
            myobject: myobject
          },
        }
      },
    }

...

有没有更好的方法来解决这个问题？

Answer 1

你必须添加它：

      reducer(state, action: PayloadAction<{x: string; anotherProp: string; uuid: string}>) {

如果您定义了一些实际上不存在的东西，这仍然会警告您：

      reducer(state, action: PayloadAction<{x: string; anotherProp: string; foo: string}>) {

但只是请求一个子类型（就像你一样）不是错误，所以它不会警告你

Redux-toolkit create action prepare 不添加 Typescript 的值

Redux-toolkit create action prepare doesn't add the values for Typescript

typescript

redux

react-redux

redux-toolkit