在房間里流動

Question

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

Pojo 用戶類

@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);
        }

依賴關系

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"

錯誤

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>)

我試圖弄清楚如何在房間里使用 rxjava，我按照示例進行操作，但它拋出了一個錯誤，有什么問題？ 可完成的工作正常

Answer 1

我不知道您為什么將ianhanniballake的答案標記為正確的答案。
依賴“ androidx.room:room-ktx:2.2.3 ”與RxJava無關。
我的情況我通過添加這個依賴項解決了這個問題

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

插入我的舊版本：

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

希望這會有所幫助

Answer 2

由於每房聲明依賴文件，你需要的依賴關系room-ktx使用協同程序，並與， Flowable ：

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