android房间关系：缺少列

Question

我在定义两个实体之间的关系时遇到问题。关系是一对一的。每个设备都分配到一个房间。

设备实体：

@Entity
data class Device(
@PrimaryKey(autoGenerate = false) val ipAddress: String,
    val deviceName: String,
    val type: Int,
    val uniqueRoomId: Int
) {
    @Ignore
    val status: DeviceStatus = DeviceStatus.Off
}

房间实体

@Entity
data class Room(
    @PrimaryKey(autoGenerate = true) val roomId: Int,
    val name: String
)

设备和房间对象：

data class DeviceAndRoom(
    @Embedded val room: Room,

    @Relation(
        parentColumn = "roomId",
        entityColumn = "uniqueRoomId",
        entity = Device::class
    )
    val device: Device
)

DeviceDao：

@Dao
interface DeviceDao {
    @Transaction
    @Query("SELECT * FROM Device")
    fun getAllDeviceAndRooms(): LiveData<List<DeviceAndRoom>>

    [...]
}

但我收到以下构建错误：

The columns returned by the query does not have the fields [roomId,name] in com.test.mytestapp.models.DeviceAndRoom even though they are annotated as non-null or primitive. Columns returned by the query: [ipAddress,deviceName,type,uniqueRoomId]

我关注了official android guidelines，但还有其他文章描述了相同的过程，例如Database relations with Room

我不知道出了什么问题。也许有人有想法。

Answer 1

对于该关系（房间（嵌入）与设备（关系）），您可以查询父级：-

@Query("SELECT * FROM room")
fun getAllDeviceAndRooms(): LiveData<List<DeviceAndRoom>>

• 名字应该是 RoomAndDevice

即它无法在设备表中找到 roomid 和 name 列，因为它们在房间表中。然后通过底层查询从每个检索到的房间中检索设备，按照SELECT * FROM the_relationship_entity WHERE entityColumn = parentColumn 的行，因此建议使用@Trasnaction。

如果您想要设备及其房间，那么您可以使用：-

data class DeviceAndRoom(
    @Embedded val device: Device,

    @Relation(
        entityColumn = "roomId",
        parentColumn = "uniqueRoomId",
        entity = Room::class
    )
    val room: Room
)

还有：-

@Transaction
@Query("SELECT * FROM Device")
fun getAllDeviceAndRooms(): LiveData<List<DeviceAndRoom>>