你可以像这样传递新值:

  setExampleState({...exampleState,  masterField2: {
        fieldOne: "a",
        fieldTwo: {
           fieldTwoOne: "b",
           fieldTwoTwo: "c"
           }
        },
   })

你可以像这样传递新值:

  setExampleState({...exampleState,  masterField2: {
        fieldOne: "a",
        fieldTwo: {
           fieldTwoOne: "b",
           fieldTwoTwo: "c"
           }
        },
   })

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" })

function App() {

  const [todos, setTodos] = useState([
    { id: 1, title: "Selectus aut autem", completed: false },
    { id: 2, title: "Luis ut nam facilis et officia qui", completed: false },
    { id: 3, title: "Fugiat veniam minus", completed: false },
    { id: 4, title: "Aet porro tempora", completed: true },
    { id: 5, title: "Laboriosam mollitia et enim quasi", completed: false }
  ]);

  const changeInput = (e) => {todos.map(items => items.id === parseInt(e.target.value) && (items.completed = e.target.checked));
 setTodos([...todos], todos);}
  return (
    <div className="container">
      {todos.map(items => {
        return (
          <div key={items.id}>
            <label>
<input type="checkbox" 
onChange={changeInput} 
value={items.id} 
checked={items.completed} />&nbsp; {items.title}</label>
          </div>
        )
      })}
    </div>
  );
}

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。

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;