可以使用useReducer钩子来管理复杂的状态，而不是useState。要做到这一点，首先初始化状态和更新函数如下所示:

const initialState = { name: "Bob", occupation: "builder" };
const [state, updateState] = useReducer(
  (state, updates) => ({ ...state, ...updates }),
  initialState
);

然后你可以通过只传递部分更新来更新状态，就像这样:

updateState({ occupation: "postman" })

2022年

如果您正在寻找与此相同的功能。函数组件中的setState(来自类组件)，那么这是对您有很大帮助的答案。

例如

你有一个像下面这样的状态，想要从整个状态中更新特定的字段，那么你需要每次都使用对象解构，有时它会令人恼火。

const [state, setState] = useState({first: 1, second: 2});

// results will be state = {first: 3} instead of {first: 3, second: 2}
setState({first: 3})

// To resolve that you need to use object destructing every time
// results will be state = {first: 3, second: 2}
setState(prev => ({...prev, first: 3}))

为了解决这个问题，我提出了useReducer方法。请检查useReducer。

const stateReducer = (state, action) => ({
  ...state,
  ...(typeof action === 'function' ? action(state) : action),
});

const [state, setState] = useReducer(stateReducer, {first: 1, second: 2});

// results will be state = {first: 3, second: 2}
setState({first: 3})

// you can also access the previous state callback if you want
// results will remain same, state = {first: 3, second: 2}
setState(prev => ({...prev, first: 3}))

您可以将该stateReducer存储在utils文件中，如果需要，也可以将其导入到每个文件中。

如果你需要，这里是自定义钩子。

import React from 'react';

export const stateReducer = (state, action) => ({
  ...state,
  ...(typeof action === 'function' ? action(state) : action),
});

const useReducer = (initial, lazyInitializer = null) => {
  const [state, setState] = React.useReducer(stateReducer, initial, init =>
    lazyInitializer ? lazyInitializer(init) : init
  );

  return [state, setState];
};

export default useReducer;

打印稿

import React, { Dispatch } from "react";

type SetStateAction<S> = S | ((prev: S) => S);

type STATE<R> = [R, Dispatch<SetStateAction<Partial<R>>>];

const stateReducer = (state, action) => ({
  ...state,
  ...(typeof action === "function" ? action(state) : action),
});

const useReducer = <S>(initial, lazyInitializer = null): STATE<S> => {
  const [state, setState] = React.useReducer(stateReducer, initial, (init) =>
    lazyInitializer ? lazyInitializer(init) : init,
  );

  return [state, setState];
};

export default useReducer;

我迟到了。：）

@aseferov回答在意图是重新进入整个对象结构时非常有效。然而，如果目标/目标是更新Object中的特定字段值，我认为下面的方法更好。

情境:

const [infoData, setInfoData] = useState({
    major: {
      name: "John Doe",
      age: "24",
      sex: "M",
    },

    minor:{
      id: 4,
      collegeRegion: "south",

    }

  });

更新特定的记录需要回调到之前的状态prevState

在这里:

setInfoData((prevState) => ({
      ...prevState,
      major: {
        ...prevState.major,
        name: "Tan Long",
      }
    }));

也许

setInfoData((prevState) => ({
      ...prevState,
      major: {
        ...prevState.major,
        name: "Tan Long",
      },
      minor: {
        ...prevState.minor,
        collegeRegion: "northEast"

    }));

我希望这对试图解决类似问题的人有所帮助。

答案已经有了，但是这种类型没有被提到，所以看看这种类型的例子…

 const[data,setdata]= useState({
    username: [
      email,
      "required", 
      //...some additional codes
    ],
    password: [
      password,
      "required|password-5",
     //..additional code if any..
    ],
  })

**要在输入字段中更新状态变量email，您可以添加类似的代码与您的变量名**

          <Input
              onChangeText={(t) => setdata(prevState=>({...prevState,username:{[0]:t}}))}
              value={data.username[0]}
            />

你想要创建状态的对象

let teams = {
  team: [
    {
      name: "one",
      id: "1"
    },
  ]
}

使团队的状态为对象

const [state, setState] = useState(teams);

像这样更新状态

setState((prevState)=>({...prevState,team:[
     ...prevState.team,
     {
     name: "two",
      id: "2"
     }
]}))

更新后状态变为

{
  team: [
    {
      name: "one",
      id: "1"
    },
    {
      name: "two",
      id: "2"
    }
  ]
}

要渲染项目根据当前状态使用地图功能

{state.team.map((curr_team) => {
      return (
        <div>
           <p>{curr_team.id}</p>
           <p>{curr_team.name}</p>
        </div>
      )
})}

您必须使用Rest参数和扩展语法(https://javascript.info/rest-parameters-spread)，并设置一个以preState作为setState参数的函数。

不起作用(功能缺失)

[state, setState] = useState({})
const key = 'foo';
const value = 'bar';
setState({
  ...state,
  [key]: value
});

确实工作!

[state, setState] = useState({})
const key = 'foo';
const value = 'bar';
setState(prevState => ({
  ...prevState,
  [key]: value
}));