我如何从React-Redux中的reducer状态更新对象数组中单个元素的名称字段?

时间:2019-12-12 12:36:56

标签: javascript reactjs redux react-redux

以下是我的减速机初始状态:

const InitialState = {
    data:[],
    SelectUser : null,
    name: "Not Yet Selected"
};

下面是从api中获取的我的data []状态:

[
 {
  "id": 1,
  "name": "Leanne Graham",
  "username": "Bret",
  "email": "Sincere@april.biz",
  "address": {
    "street": "Kulas Light",
    "suite": "Apt. 556",
    "city": "Gwenborough",
    "zipcode": "92998-3874",
    "geo": {
      "lat": "-37.3159",
      "lng": "81.1496"
    }
  },
  "phone": "1-770-736-8031 x56442",
  "website": "hildegard.org",
  "company": {
    "name": "Romaguera-Crona",
    "catchPhrase": "Multi-layered client-server neural-net",
    "bs": "harness real-time e-markets"
  }
 },
 {
  "id": 2,
  "name": "Ervin Howell",
  "username": "Antonette",
  "email": "Shanna@melissa.tv",
  "address": {
    "street": "Victor Plains",
    "suite": "Suite 879",
    "city": "Wisokyburgh",
    "zipcode": "90566-7771",
    "geo": {
      "lat": "-43.9509",
      "lng": "-34.4618"
    }
  },
  "phone": "010-692-6593 x09125",
  "website": "anastasia.net",
  "company": {
    "name": "Deckow-Crist",
    "catchPhrase": "Proactive didactic contingency",
    "bs": "synergize scalable supply-chains"
  }
 }
]

我想将名称从状态为 data [] 的对象更改为状态 name 。其中,对象的id与 SelecUser 状态的对象中的id相同。 SelectUser 也是对象,其内容与 data [] 中的元素相同。所有状态都已设置。

以下是迄今为止创建的减速器:


const InitialState = {
    data:[],
    SelectUser : null,
    name: "Not Yet Selected"
};

export default (state=InitialState, action)=>{
    switch(action.type){
            case 'FETCH_USER':
                return {...state,data:action.payload};
            case 'USER_SELECTED':
                return {...state,SelectUser:action.payload};
            case 'USER_CHANGE':
                return {...state,name:action.payload};
            default:
                return state;
    }
};

所有提到的情况以前都已执行。

2 个答案:

答案 0 :(得分:1)

  

我想将名称从状态数据[]中的对象更改为状态名称   具有相同ID的对象(已存储在SelectUser中)

我假设您要更改name的{​​{1}}上端,为简单起见,我将其称为action

CHANGE_NAME

现在,您只需要在数据中找到对应对象case 'CHANGE_NAME' return handleNameChange(state, action) 是原始内容并覆盖spread

name

答案 1 :(得分:0)

尝试一下

export default (state=InitialState, action)=>{
    switch(action.type){
            case 'FETCH_USER':
                return {...state,data:action.payload};
            case 'USER_SELECTED':
                return {...state,SelectUser:action.payload};
            case 'USER_CHANGE':
                return {
                         ...state,
                         name: state.data.reduce((prev, item)=>{
                                  if(state.SelectUser && state.SelectUser.some(elt=>elt.id === item.id)){
                                      return item.name;
                                  }else{
                                      return prev;
                                  }
                         }, null);
                };
            default:
                return state;
    }
};