Paging3："Not sure how to convert a Cursor to this method's return type" 将 PagingSource 用作 Room DAO 中的 return 类型

Question

我试图为新的 Paging 3 库模仿 Google 的代码实验室，当我尝试使用 Room DAO 方法 return a PagingSource:

D:\Programming\Android\something\app\build\tmp\kapt3\stubs\debug\com\someapp\something\data\db\UsersDao.java:38: error: Not sure how to convert a Cursor to this method's return type (androidx.paging.PagingSource<java.lang.Integer,com.someapp.something.data.db.GithubUser>).
    public abstract androidx.paging.PagingSource<java.lang.Integer, com.someapp.something.data.db.GithubUser> getUserByUserName(@org.jetbrains.annotations.NotNull()
    
^D:\Programming\Android\something\app\build\tmp\kapt3\stubs\debug\com\someapp\something\data\db\UsersDao.java:43: error: Not sure how to convert a Cursor to this method's return type (androidx.paging.PagingSource<java.lang.Integer,com.someapp.something.data.db.GithubUser>).
public abstract androidx.paging.PagingSource<java.lang.Integer, com.someapp.something.data.db.GithubUser> getUserByNote(@org.jetbrains.annotations.NotNull()

这是我的 UsersDao.kt:

@Dao
interface UsersDao {

    @Insert
    fun insert(user: GithubUser): Completable

    @Insert
    fun insert(userList: List<GithubUser>): Completable

    @Query("DELETE FROM userDb")
    fun clearDb(): Completable

    @Query("SELECT * FROM userDb")
    fun getAllUsers(): Single<List<GithubUser>>

    @Query("SELECT EXISTS(SELECT 1 FROM userDb WHERE username LIKE :userName)")
    fun checkIfUserExists(userName: String): Boolean

    @Query("SELECT note FROM userDb WHERE username LIKE :userName")
    fun getNoteByUserName(userName: String): Single<String>

    @Query("SELECT * FROM userDb WHERE username LIKE :userName")
    fun getUserByUserName(userName: String): PagingSource<Int, GithubUser>

    @Query("SELECT * FROM userDb WHERE note LIKE :note")
    fun getUserByNote(note: String): PagingSource<Int, GithubUser>

}

我的 GithubUser.kt 看起来像这样：

@Entity(tableName = "userDb", indices = arrayOf(Index(value = ["username"],  unique = true)))
class GithubUser (
    var username: String,
    var note: String,
    var url: String,
    var avatarUrl: String
) {
    @PrimaryKey(autoGenerate = true)
    var uid = 0
}

在使用 Paging 3 库的 code for the Paging Codelab, the DAO method just returns a PagingSource with no extra annotations/magic options in Gradle or whatever. I also looked at other examples from Github like this and this 中，它们 return PagingSource 完全没有问题。谁能告诉我是否遗漏了什么？

注意：在错误本身之前，我总是收到关于 ANTLR Tool version 4.5.3 used for code generation does not match the current runtime version 4.7.1 的警告，但这个警告本身在过去并没有引起任何问题，但我以防万一。

编辑：我使用以下 Room/Paging 库版本：

    implementation "androidx.room:room-runtime:2.2.5"
    kapt "androidx.room:room-compiler:2.2.5"
    implementation 'androidx.room:room-rxjava2:2.2.5'

    implementation "androidx.paging:paging-runtime:3.0.0-alpha03"
    implementation 'androidx.paging:paging-rxjava2:3.0.0-alpha03'

Answer 1

原来需要把Room版本提高到2.3.0-alpha02以上：

implementation "androidx.room:room-runtime:2.3.0-alpha02"
implementation "androidx.room:room-ktx:2.3.0-alpha02"
kapt "androidx.room:room-compiler:2.3.0-alpha02"

Paging3："Not sure how to convert a Cursor to this method's return type" 将 PagingSource 用作 Room DAO 中的 return 类型

Paging3: "Not sure how to convert a Cursor to this method's return type" when using PagingSource as return type in Room DAO

android

android-room

rx-java2

android-paging

android-paging-library