Question

我有一个展示名称和状态的Flatlist，我想通过单击平面列表项来切换用户的状态。我已经为此创建了redux动作和redux减速器。但是无法改变相应的状态。

{
"response": [{
  "name": "RAJ",
   "status": true

 }, {
 "name": "RAM",
  "status": true,

}]
}


  <FlatList
     showsVerticalScrollIndicator={false}
     removeClippedSubviews={false}
     data={this.props.response}
     ItemSeparatorComponent = {this.flatListItemSeparator}
     ListFooterComponent={this.flatListItemSeparator}
     keyExtractor={(item, index) => index}
     renderItem={({item,index}) => this.renderFlatListItem(item,index)}/>

但无法更改与指定用户对应的状态。例如Raj状态为false

{{1}}

Answer 1

我不知道这是否正是你想要实现的目标，但我会试一试。

//call Action 
this.props.listItemClick('RAJ',false});

//Action
export const listItemClick = (user,status) => {
    return {
        type: LIST_CLICK,
        payload: { user, status }
    };
}; 

 export default (state = INITIAL_STATE, action) => {
     switch (action.type) {
        case LIST_CLICK:
            //replace YOUR_DATA with your state var 
            return {...state, YOUR_DATA: [state.YOUR_DATA.map((val,index) => {
                   if (val.name === action.payload.user) {
                      return { ...val, status: action.payload.status };
                   }
                   return { ...val };
            })]};
        default:
            return state;

    }
};

如何使用redux（Flatlist项目单击）更新特定json数组项中的单个值？

1 个答案: