我给您留下了一个实用函数来不可变地更新对象

/**
 * Inmutable update object
 * @param  {Object} oldObject     Object to update
 * @param  {Object} updatedValues Object with new values
 * @return {Object}               New Object with updated values
 */
export const updateObject = (oldObject, updatedValues) => {
  return {
    ...oldObject,
    ...updatedValues
  };
};

你可以这样用

const MyComponent = props => {

  const [orderForm, setOrderForm] = useState({
    specialities: {
      elementType: "select",
      elementConfig: {
        options: [],
        label: "Specialities"
      },
      touched: false
    }
  });


// I want to update the options list, to fill a select element

  // ---------- Update with fetched elements ---------- //

  const updateSpecialitiesData = data => {
    // Inmutably update elementConfig object. i.e label field is not modified
    const updatedOptions = updateObject(
      orderForm[formElementKey]["elementConfig"],
      {
        options: data
      }
    );
    // Inmutably update the relevant element.
    const updatedFormElement = updateObject(orderForm[formElementKey], {
      touched: true,
      elementConfig: updatedOptions
    });
    // Inmutably update the relevant element in the state.
    const orderFormUpdated = updateObject(orderForm, {
      [formElementKey]: updatedFormElement
    });
    setOrderForm(orderFormUpdated);
  };

  useEffect(() => {
      // some code to fetch data
      updateSpecialitiesData.current("specialities",fetchedData);
  }, [updateSpecialitiesData]);

// More component code
}

如果没有，您可以在这里找到更多实用程序:https://es.reactjs.org/docs/update.html

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" })

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} 
/>

最初我在useState中使用object，但后来我移动到useReducer钩子用于复杂的情况。重构代码时，我感到性能有所提高。

当您有涉及多个子值的复杂状态逻辑时，或者当下一个状态依赖于前一个状态时，useReducer通常比useState更可取。

useReducer React文档

我已经实现了这样的钩子供我自己使用:

/**
 * Same as useObjectState but uses useReducer instead of useState
 *  (better performance for complex cases)
 * @param {*} PropsWithDefaultValues object with all needed props 
 * and their initial value
 * @returns [state, setProp] state - the state object, setProp - dispatch 
 * changes one (given prop name & prop value) or multiple props (given an 
 * object { prop: value, ...}) in object state
 */
export function useObjectReducer(PropsWithDefaultValues) {
  const [state, dispatch] = useReducer(reducer, PropsWithDefaultValues);

  //newFieldsVal={[field_name]: [field_value], ...}
  function reducer(state, newFieldsVal) {
    return { ...state, ...newFieldsVal };
  }

  return [
    state,
    (newFieldsVal, newVal) => {
      if (typeof newVal !== "undefined") {
        const tmp = {};
        tmp[newFieldsVal] = newVal;
        dispatch(tmp);
      } else {
        dispatch(newFieldsVal);
      }
    },
  ];
}

更多相关的钩子。

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;