android - 如何为另一个 ViewModel 中的对象属性派生 ViewModel？

问题描述

我想为另一个 ViewModel 中保存的另一个对象的属性获取一个 ViewModel。

我有这样的关系：在一个房子里有很多人。（人在 Houses 表中编码的 1:n 关系，而不是使用连接表。）在这种情况下我有一个问题：现有的 House 将显示在 HouseDetailsActivity 中，其中包含一个 HouseDetailsFragment 和一个 PeopleListFragment。HouseDetailsActivity 在 onCreate 中获取 HouseViewModel，如下所示：

houseViewModel = ViewModelProviders.of(this, new HouseViewModel.Factory(getApplication(), id)).get(HouseViewModel.class);

HouseViewModel 能够返回 LiveData，因为它从数据库中获取了 HouseEntity。PeopleListFragment 需要从某个地方获取该房子的人员列表的 LiveData，但不需要了解 PeopleListViewModel 以外的任何视图模型。所以，同样在 HouseDetailsActivity onCreate 中，我得到了一个 PeopleListViewModel，如下所示：

peopleListViewModel = ViewModelProviders.of(this).get(PeopleListViewModel.class);

我希望可以与 PeopleListFragment 共享，如下所示：

peopleViewModel = ViewModelProviders.of(getActivity()).get(PeopleListViewModel.class);

问题是如何将 LiveData 中的人员列表放入 ViewModel。HouseDetailsActivity (HouseDetailsViewModel) 内的 HouseEntity 中的人员列表不是 LiveData。（我希望能够通过 PeopleListViewModel 在 PeopleListFragment 中查看 HouseEntity 的人员列表。）

我已经看过 MediatorLiveData 的文档，我认为这里不适用，因为最终只有一个 PeopleList 来源。

public class HouseDetailsActivity
{
protected void onCreate(Bundle savedInstanceState)
{
    houseViewModel = ViewModelProviders.of(this, new HouseViewModel.Factory(getApplication(), id)).get(HouseViewModel.class);
    peopleListViewModel = ViewModelProviders.of(this).get(PeopleListViewModel.class);
    /* This can't be done, because the HouseEntity may not yet be loaded to the ViewModel. ie. NullPointerException here
    List<Person> people = m_houseViewModel.getHouse().getPeopleList();
    peopleListViewModel.setPeople(people);
    */
}
}

@Entity(tableName="houses")
public class HouseEntity implements MutableHouse
{
    @PrimaryKey(autoGenerate = true)
    @ColumnInfo(name="hid")
    public int id = 0;

    @ColumnInfo(name="address")
    private String address = null;

    /** This is the encoded people, for multiple in a single database field. */
    @ColumnInfo(name="residents")
    private String residents = null;

    public List<Person> getPeopleList ()
    { return HouseEncoding.decodePeople(getResidents()); }
    ...
}

public class HouseViewModel
{
    private final int houseId;
    private MutableLiveData<HouseEntity> house; // The list of people is inside house here, but not as LiveData

    public LiveData<HouseEntity> getObservableHouse ()
    { return house; }

    HouseViewModel (@NonNull Application application, int houseId)
    {
        super(application);
        this.houseId = houseId;
        this.house = getRepository().getHouseObservable(houseId);
    }

    /**
     * A creator is used to inject the house ID into the ViewModel
     */
    public static class Factory extends ViewModelProvider.NewInstanceFactory
    {
        @NonNull
        private Application application;
        private int houseId;

        public Factory (@NonNull Application application, int houseId)
        {
            this.application = application;
            this.houseId = houseId;
        }

        @Override
        @NonNull
        public <T extends ViewModel> T create (@NonNull Class<T> modelClass)
        {
            //noinspection unchecked
            return (T) new HouseViewModel(application, houseId);
        }
    }
}

public class PeopleListViewModel
{
    private MutableLiveData<List<Person>> people;

    void setPeople (List<Person> people)
    { this.people.setValue(people); }
    ...
}

在 PeopleListFragment 中：

private void observerSetup ()
{
    peopleViewModel.getPeople().observe(this, people -> {
        adapter.setPeople(people); // for RecyclerView
    });
}

标签： androidviewmodelandroid-livedataandroid-viewmodel

android - 如何为另一个 ViewModel 中的对象属性派生 ViewModel？

问题描述

解决方案

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