Question

我的redux状态看起来像这样（与Firebase同步）。

@IBAction func settingsExitButtonPressed(_ sender: UIBarButtonItem) {

    dismiss(animated: true, completion: nil)

//  This is used to let the Root View know to only reload itself 
//  when coming from the settings.
    didOpenFromSettings = true


}

我要删除用户zQiGXvcUGmRSKUdr719621QleUw2

这里是我的动作创建者

{
  profile: {
    activeUsers: {
      Iiva2BGZffNTH84glOLXv8QHVTF2: {
        sex: male,
        age: 20,
      },
      PkfMxrN09RN7ygoBMWqm4jheEOx1: {
        sex: female,
        age: 20,
      },
      zQiGXvcUGmRSKUdr719621QleUw2: {
        sex: male,
        age: 25,
      }
    }
  }
}

最后是我的减速器

  Firebase.database()
    .ref('profiles/activeUsers')
    .on(
      'child_removed',
      (snapshot) => {
        dispatch(_activeUserChildRemoved(snapshot));
      },
      (err) => {
        console.log(err.toString());
        Alert.alert(err);
      },
    );
};

const _activeUserChildRemoved = snapshot => ({
  type: ACTIVE_USER_CHILD_REMOVED,
  payload: snapshot,
});

为了从redux中删除snapshot.key引用的用户，我应该从reducer返回什么？非常感谢帮助

Answer 1

知道了！！

case ACTIVE_USER_CHILD_REMOVED:
      const key4Del = action.payload.key;
      const oldState = state;
      delete oldState.activeUsers[key4Del];
      return { ...oldState };

Answer 2

您需要通过不进行变异状态，而只需使用解构来删除用户。

现在，您要通过从state删除对象属性来直接对oldState进行突变

case ACTIVE_USER_CHILD_REMOVED:
  const key4Del = action.payload.key;
  const { [key4Del]: _, ...activeUsers} = state.activeUsers
  return {...state, activeUsers }

“扩展语法”将浅比较对象，因此您需要再次合并已过滤的对象以生成新状态。

在触发Firebase'child_removed'侦听器后，将嵌套的子级从redux状态中删除

2 个答案: