Android Room database@Insert方法返回从0开始的行id

Question

我已经在我的应用程序中实现了MVVM架构。我有我的活动、视图模型、存储库、DAO和数据库类。数据库保存包含不同列表的对象，我想在这些列表之间切换，并让我的RecycraView显示当前选定的列表。我的理解是SQLite中的行ID从1开始，但是我的存储库中的插入方法（在我的DAO中调用@插入方法）总是从0开始。出于测试目的，我使用LiveData获取数据库中的所有对象，当我记录它们的ID时，它们正确地从1开始，一直到n。我不想在内存中维护数据库中所有对象的列表，只维护包含当前选定列表的单个对象。当用户创建一个新列表（或选择一个现有列表）时，我希望我的回收站视图显示其内容，并观察该列表的任何更改。没有相应的ID，我无法开始观察对象。

如何将正确的ID传播到MainActivity？如果需要我的ViewModel和活动代码，请告诉我，我将编辑此帖子。

当我从存储库中记录ID时

当我在MainActivity中记录LiveData时，会显示正确的ID

我的数据库类

@TypeConverters({ArrayListConverter.class}) // List converter for database
@Database(entities = ListContainer.class, version = 1, exportSchema = false)
public abstract class ListContainerDatabase extends RoomDatabase {

  private static ListContainerDatabase dbInstance;
  private static final String DB_NAME = "list_container_db";
  private static final int NUM_THREADS = 4;

  final static ExecutorService executor = Executors.newFixedThreadPool(NUM_THREADS);

  public abstract ListContainerDao getDao();

  public static synchronized ListContainerDatabase getInstance(Context context) {
    if(dbInstance == null) {
      dbInstance = Room.databaseBuilder(
              context.getApplicationContext(),
              ListContainerDatabase.class, DB_NAME)
              .fallbackToDestructiveMigration()
              .build();
    }
    return dbInstance;
  }

  public static Executor getExecutor() {
    return executor;
}

我的DAO

@Dao
public interface ListContainerDao {

  @Insert
  long insertListContainer(ListContainer container);

... // other database queries

}

我的资源库

public class Repository {

  private static final String TAG = "Repository";

  private ListContainerDao listContainerDao;
  private long id;

  private Executor executor = ListContainerDatabase.getExecutor();

  public Repository(Application application) {
    ListContainerDatabase containerDb = ListContainerDatabase.getInstance(application);
    listContainerDao = containerDb.getDao();
  }

  public long insertListContainer(ListContainer container) {
    executor.execute(new Runnable() {
      @Override
      public void run() {
        id = listContainerDao.insertListContainer(container);  // this returns ID starting at 0
      }
    });
    Log.i(TAG, "insertListContainer: from Repository id is :" + id);
    return id;
  }
}

Answer 1

您应该在这里考虑代码的异步性：

private long id;

public long insertListContainer(ListContainer container) {
    // Timestamp #1. This started, first time id is not initialised, equals to 0
    executor.execute(new Runnable() { 
    @Override
      public void run() {
        id = listContainerDao.insertListContainer(container);  
        // Timestamp #3. This returns you id = 1, but with some delay, so this value changed id and will be returned only next time you call the method 
      }
    });
    // Timestamp #2. Main thread isn't blocked, so it doesn't wait for runnable to be executed as well and returns first time id = 0, next time - changed value - 1, and so on
    Log.i(TAG, "insertListContainer: from Repository id is :" + id); 
    return id;
  }

如果你想在Java得到异步操作的结果，你可以使用：

• 回调（类似于此答案）
• RxJava/CompletableFuture