一般来说，你应该注意React状态下嵌套很深的对象。为了避免意外的行为，状态应该不可更改地更新。当你有深层对象时，你最终会为了不变性而对它们进行深层克隆，这在React中是相当昂贵的。为什么?

一旦你深度克隆状态，React将重新计算和重新渲染所有依赖于变量的东西，即使它们没有改变!

因此，在尝试解决问题之前，首先考虑如何将状态变平。一旦您这样做了，您就会发现有助于处理大型状态的方便工具，例如useReducer()。

如果你想过，但仍然确信你需要使用深度嵌套的状态树，你仍然可以使用useState()与immutable.js和Immutability-helper等库一起使用。它们使得更新或克隆深层对象变得简单，而不必担心可变性。

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
}));

谢谢Philip，这帮助了我-我的用例是我有一个有很多输入字段的表单，所以我保持初始状态为对象，我不能更新对象状态。上面的帖子帮助了我:)

const [projectGroupDetails, setProjectGroupDetails] = useState({
    "projectGroupId": "",
    "projectGroup": "DDD",
    "project-id": "",
    "appd-ui": "",
    "appd-node": ""    
});

const inputGroupChangeHandler = (event) => {
    setProjectGroupDetails((prevState) => ({
       ...prevState,
       [event.target.id]: event.target.value
    }));
}

<Input 
    id="projectGroupId" 
    labelText="Project Group Id" 
    value={projectGroupDetails.projectGroupId} 
    onChange={inputGroupChangeHandler} 
/>

如果你使用布尔值和数组，这可以帮助你:

const [checkedOrders, setCheckedOrders] = useState<Record<string, TEntity>>({});

const handleToggleCheck = (entity: TEntity) => {
  const _checkedOrders = { ...checkedOrders };
  const isChecked = entity.id in checkedOrders;

  if (isChecked) {
    delete _checkedOrders[entity.id];
  } else {
    _checkedOrders[entity.id] = entity;
  }

  setCheckedOrders(_checkedOrders);
};

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]}
            />

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;