Android Room - Get the id of new inserted row with auto-generate

问题

This is how I am inserting data into database using Room Persistence Library:

Entity:

@Entity
class User {
    @PrimaryKey(autoGenerate = true)
    public int id;
    //...
}

Data access object:

@Dao
public interface UserDao{
    @Insert(onConflict = IGNORE)
    void insertUser(User user);
    //...
}

Is it possible to return the id of User once the insertion is completed in the above method itself without writing a separate select query?

回答1:

Based on the documentation here (below the code snippet)

A method annotated with the @Insert annotation can return:

long for single insert operation

long[] or Long[] or List<Long> for multiple insert operations

void if you don't care about the inserted id(s)

回答2:

@Insert function can return void, long, long[] or List<Long>. Please try this.

 @Insert(onConflict = OnConflictStrategy.REPLACE)
  long insert(User user);

 // Insert multiple items
 @Insert(onConflict = OnConflictStrategy.REPLACE)
  long[] insert(User... user);

回答3:

The return value of the insertion for one record will be 1 if your statement successfully.

In case you want to insert list of objects, you can go with:

@Insert(onConflict = OnConflictStrategy.REPLACE)
public long[] addAll(List<Object> list);

And execute it with Rx2:

Observable.fromCallable(new Callable<Object>() {
        @Override
        public Object call() throws Exception {
            return yourDao.addAll(list<Object>);
        }
    }).subscribeOn(Schedulers.io()).observeOn(AndroidSchedulers.mainThread()).subscribe(new Consumer<Object>() {
        @Override
        public void accept(@NonNull Object o) throws Exception {
           // the o will be Long[].size => numbers of inserted records.

        }
    });

回答4:

Get the row ID by the following sniplet. It uses callable on an ExecutorService with Future.

 private UserDao userDao;
 private ExecutorService executorService;

 public long insertUploadStatus(User user) {
    Callable<Long> insertCallable = () -> userDao.insert(user);
    long rowId = 0;

    Future<Long> future = executorService.submit(insertCallable);
     try {
         rowId = future.get();
    } catch (InterruptedException e1) {
        e1.printStackTrace();
    } catch (ExecutionException e) {
        e.printStackTrace();
    }
    return rowId;
 }

Ref: Java Executor Service Tutorial for more information on Callable.

回答5:

In you Dao, the query insert return Long i.e. the rowId inserted.

 @Insert(onConflict = OnConflictStrategy.REPLACE)
 fun insert(recipes: CookingRecipes): Long

In your Model(Repository) class : (MVVM)

fun addRecipesData(cookingRecipes: CookingRecipes): Single<Long>? {
        return Single.fromCallable<Long> { recipesDao.insertManual(cookingRecipes) }
}

In your ModelView class: (MVVM) Handle LiveData with DisposableSingleObserver.
Working sourcer reference : https://github.com/SupriyaNaveen/CookingRecipes

来源：https://stackoverflow.com/questions/44364240/android-room-get-the-id-of-new-inserted-row-with-auto-generate

标签

android

persistent-storage

android-room