反应本机状态管理问题-useState挂钩何时加载?

时间:2019-12-17 19:02:37

标签: react-native state react-hooks react-state-management react-state

我有一个项目的平面清单,旁边有一个“删除”按钮。

当我单击删除按钮时,我能够从后端删除该项目,但实际的列表项目并未从视图中删除。

我正在使用useState挂钩,据我了解,组件在setState发生后会重新渲染。

  

setState函数用于更新状态。它接受一个新的   状态值,并重新渲染组件。   https://reactjs.org/docs/hooks-reference.html

状态设置和呈现方式我缺少什么?

出于各种原因,我不想使用useEffect侦听器。我希望组件在位置状态更新时重新呈现....我很确定正在与其他setStates一起发生....不确定我是否完全缺少setState所做的标记或是否这与setLocations()有关。

    const [locations, setLocations] = useState(state.infoData.locations);
    const [locationsNames, setLocationsNames] = useState(state.infoData.names]);

    ...

    const removeLocationItemFromList = (item) => {
      var newLocationsArray = locations;
      var newLocationNameArray = locationsNames;
      for(l in locations){
        if(locations[l].name == item){
          newLocationsArray.splice(l, 1);
          newLocationNameArray.splice(l, 1);
        } else {
          console.log('false');
        }
      }
      setLocationsNames(newLocationNameArray); 
      setLocations(newLocationsArray);
    };

    ...

    <FlatList style={{borderColor: 'black', fontSize: 16}} 
      data={locationNames} 
      renderItem={({ item }) => 
          <LocationItem
            onRemove={() => removeLocationItemFromList(item)}
            title={item}/> } 
      keyExtractor={item => item}/>

更新后的循环

const removeLocationItemFromList = (item) => {
  var spliceNewLocationArray =locations;
  var spliceNewLocationNameArray = locationsNames;
  for(f in spliceNewLocationArray){
    if(spliceNewLocationArray[f].name == item){
      spliceNewLocationArray.splice(f, 1);
    } else {
      console.log('false');
    }
  }
  for(f in spliceNewLocationNameArray){
    if(spliceNewLocationNameArray[f] == item){
      spliceNewLocationNameArray.splice(f, 1);
    } else {
      console.log('false');
    }
  }
  var thirdTimesACharmName = spliceNewLocationNameArray;
  var thirdTimesACharmLoc = spliceNewLocationArray;
  console.log('thirdTimesACharmName:: ' + thirdTimesACharmName + ', thirdTimesACharmLoc::: ' + JSON.stringify(thirdTimesACharmLoc)); // I can see from this log that the data is correct
  setLocationsNames(thirdTimesACharmName); 
  setLocations(thirdTimesACharmLoc);
};

1 个答案:

答案 0 :(得分:1)

这归结为更改相同的locations数组并再次调用同一数组的setState,这意味着作为纯组件的FlatList将不会重新呈现,因为locations的标识未更改。您可以先将locations数组复制到newLocationsArray(与newLocationNameArray相似)来避免这种情况。

      var newLocationsArray = locations.slice();
      var newLocationNameArray = locationsNames.slice();