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

我认为更优雅的解决方案是创建更新后的状态对象,同时保留以前的state值。需要更新的Object属性可以以这样的数组形式提供

import React,{useState, useEffect} from 'react'
export default function Home2(props) {
    const [x, setX] = useState({name : '',add : {full : '', pin : '', d : { v : '' }}})
    const handleClick = (e, type)=>{
        let obj = {}
        if(type.length > 1){
            var z = {}
            var z2 = x[type[0]]
        
        type.forEach((val, idx)=>{
            if(idx === type.length - 1){
                z[val] = e.target.value
            }
            else if(idx > 0){
                Object.assign(z , z2) /*{...z2 , [val]:{} }*/
                z[val] = {}
                z = z[val]
                z2 = z2[val]
            }else{
                z = {...z2}
                obj = z
            }
        })
    }else obj = e.target.value
    setX( { ...x ,   [type[0]] : obj  } )
    
}
return (
    <div>
        <input value = {x.name} onChange={e=>handleClick(e,["name"])}/>
        <input value = {x.add.full} onChange={e=>handleClick(e,["add","full"])}  />
        <input value = {x.add.pin} onChange={e=>handleClick(e,["add","pin"])}  /><br/>
        <input value = {x.add.d.v} onChange={e=>handleClick(e,["add","d","v"])}  /><br/>
        {x.name} <br/>
        {x.add.full} <br/>
        {x.add.pin} <br/>
        {x.add.d.v}
    </div>
)
}

其他回答

我认为更优雅的解决方案是创建更新后的状态对象,同时保留以前的state值。需要更新的Object属性可以以这样的数组形式提供

import React,{useState, useEffect} from 'react'
export default function Home2(props) {
    const [x, setX] = useState({name : '',add : {full : '', pin : '', d : { v : '' }}})
    const handleClick = (e, type)=>{
        let obj = {}
        if(type.length > 1){
            var z = {}
            var z2 = x[type[0]]
        
        type.forEach((val, idx)=>{
            if(idx === type.length - 1){
                z[val] = e.target.value
            }
            else if(idx > 0){
                Object.assign(z , z2) /*{...z2 , [val]:{} }*/
                z[val] = {}
                z = z[val]
                z2 = z2[val]
            }else{
                z = {...z2}
                obj = z
            }
        })
    }else obj = e.target.value
    setX( { ...x ,   [type[0]] : obj  } )
    
}
return (
    <div>
        <input value = {x.name} onChange={e=>handleClick(e,["name"])}/>
        <input value = {x.add.full} onChange={e=>handleClick(e,["add","full"])}  />
        <input value = {x.add.pin} onChange={e=>handleClick(e,["add","pin"])}  /><br/>
        <input value = {x.add.d.v} onChange={e=>handleClick(e,["add","d","v"])}  /><br/>
        {x.name} <br/>
        {x.add.full} <br/>
        {x.add.pin} <br/>
        {x.add.d.v}
    </div>
)
}

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>
      )
})}

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。

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]}
            />
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>
  );
}