在嵌套对象中,在React with Hooks中更新状态的正确方法是什么?

export Example = () => {
  const [exampleState, setExampleState] = useState(
  {masterField: {
        fieldOne: "a",
        fieldTwo: {
           fieldTwoOne: "b"
           fieldTwoTwo: "c"
           }
        }
   })

如何使用setExampleState将exampleState更新为a(附加字段)?

const a = {
masterField: {
        fieldOne: "a",
        fieldTwo: {
           fieldTwoOne: "b",
           fieldTwoTwo: "c"
           }
        },
  masterField2: {
        fieldOne: "c",
        fieldTwo: {
           fieldTwoOne: "d",
           fieldTwoTwo: "e"
           }
        },
   }
}

b(改变值)?

const b = {masterField: {
        fieldOne: "e",
        fieldTwo: {
           fieldTwoOne: "f"
           fieldTwoTwo: "g"
           }
        }
   })

当前回答

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);
};

其他回答

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

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。

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。

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。

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>
  );
}