Question

我遇到以下错误：

error: Type of the parameter must be a class annotated with @Entity or a collection/array of it.
    java.lang.String matchId);

在我的matchedBefore()查询中：

@Dao
interface MatchedUsersDao {

    // Checks to see if user has matched before (if it returns 1)
    @Query("SELECT COUNT(*) FROM matched_users WHERE :matchId = match_id LIMIT 1")
    fun matchedBefore(matchId: String): Int

    @Insert(onConflict = OnConflictStrategy.ABORT)
    fun addMatchUid(matchId: String)
}

这是我的实体

@Entity(tableName = "matched_users")
data class MatchedUser(
    @PrimaryKey(autoGenerate = true) val id: Int,
    @ColumnInfo(name = "match_id") val matchId: String
)

数据库

@Database(entities = arrayOf(MatchedUser::class), version = 1)
abstract class AppDatabase : RoomDatabase() {

    abstract fun matchedUsersDao(): MatchedUsersDao
}

然后我在应用程序中实例化数据库：

class CustomApplication : Application() {

    companion object {
        var database: AppDatabase? = null
    }

    override fun onCreate() {
        super.onCreate()
        CustomApplication.database = Room.databaseBuilder(this, AppDatabase::class.java, "AppDatabase").build()
    }

有人可以告诉我该错误指示什么以及如何解决吗？

Answer 1

您的问题是这行

@Insert(onConflict = OnConflictStrategy.ABORT)
fun addMatchUid(matchId: String)

如果要使用@Insert批注，则必须发送MatchedUser而不是String的实例。

因此您可以：

1）发送具有您的ID的新MatchedUser（插入具有您的ID的空MatchedUser）

您的代码必须是这样

@Insert(onConflict = OnConflictStrategy.ABORT)
fun addMatchUid(matchUser: MatchedUser)

2）编写了新的@Query以便仅在数据库中插入ID

房间错误：参数的类型必须是带有@Entity或其集合/数组的类

1 个答案: