可在房间内流动

Question

@Database(entities = {User.class}, version = 2, exportSchema = false)
public abstract class AppDatabase extends RoomDatabase {
    public abstract userDao userDao();
}

Pojo 用户class

@Entity
    public class User {
        @PrimaryKey(autoGenerate = true)
        private int id;

        public User(){
        }



        public int getId() {
            return id;
        }

        public void setId(int id) {
            this.id = id;
        }
    }

道

@Dao
public interface userDao {
            @Query("SELECT * FROM User WHERE id = :id")
            Flowable<User> get(int id);
            @Insert
            Completable insert(User user);
        }

依赖项

implementation "androidx.room:room-runtime:2.2.3"
annotationProcessor "androidx.room:room-compiler:2.2.3"
implementation "android.arch.persistence.room:rxjava2:1.1.1"
implementation 'io.reactivex.rxjava2:rxandroid:2.1.1'
implementation "io.reactivex.rxjava2:rxjava:2.2.14"

错误

error: no suitable method found for createFlowable(RoomDatabase,boolean,String[],<anonymous Callable<User>>)
method RxRoom.createFlowable(RoomDatabase,String...) is not applicable
(varargs mismatch; boolean cannot be converted to String)
method RxRoom.<T>createFlowable(RoomDatabase,String[],Callable<T>) is not applicable
(cannot infer type-variable(s) T
(actual and formal argument lists differ in length))
where T is a type-variable:
T extends Object declared in method <T>createFlowable(RoomDatabase,String[],Callable<T>)

我正在尝试弄清楚如何在房间中使用 rxjava，我按照示例进行操作，但它会引发错误，问题是什么？可完成工作正常

Answer 1

根据 Room Declaring Dependencies documentation，您需要依赖 room-ktx 才能使用协程，并且 Flowable:

implementation "androidx.room:room-ktx:2.2.3"

Answer 2

我不知道你为什么将 ianhanniballake 的答案标记为正确答案。
依赖项“androidx.room:room-ktx:2.2.3”与 RxJava 无关。
我的情况是我通过添加此依赖项

解决了问题

implementation "androidx.room:room-rxjava2:2.2.3"

代替我的旧版本：

implementation 'android.arch.persistence.room:rxjava2:1.1.1'

希望这会有所帮助

可在房间内流动

Flowable in room

java

android

rx-java

android-room