在列名相同的情况下，如何表示Android Room的多对多关系？
如何用Room表示“多对多”关系？我的列名也是一样的。
例如，我有Guest和Reservation。Reservation可以有多个Guest，一个Guest可以是许多预留的一部分。
以下是我的实体定义：
@Entity data class Reservation(
@PrimaryKey val id: Long,
val table: String,
val guests: List<Guest>
)
@Entity data class Guest(
@PrimaryKey val id: Long,
val name: String,
val email: String
)在查看文档时，我遇到了@Relation。不过，我发现它真的很令人困惑。
根据这一点，我想创建一个POJO，并在那里添加关系。因此，在我的示例中，我执行了以下操作：
data class ReservationForGuest(
@Embedded val reservation: Reservation,
@Relation(
parentColumn = "reservation.id",
entityColumn = "id",
entity = Guest::class
) val guestList: List<Guest>
)在上面的代码中，我得到了编译器错误：
> Cannot figure out how to read this field from a cursor.我找不到@Relation的工作样本。
回答 5
Stack Overflow用户
发布于 2017-06-08 14:42:22
我也遇到过类似的问题。这是我的解决方案。
您可以使用一个额外的实体(ReservationGuest)来保持Guest和Reservation之间的关系。
@Entity data class Guest(
@PrimaryKey val id: Long,
val name: String,
val email: String
)
@Entity data class Reservation(
@PrimaryKey val id: Long,
val table: String
)
@Entity data class ReservationGuest(
@PrimaryKey(autoGenerate = true) val id: Long,
val reservationId: Long,
val guestId: Long
)您可以通过他们的guestId列表(而不是来宾对象)获得预订。
data class ReservationWithGuests(
@Embedded val reservation:Reservation,
@Relation(
parentColumn = "id",
entityColumn = "reservationId",
entity = ReservationGuest::class,
projection = "guestId"
) val guestIdList: List<Long>
)您还可以使用客户的预订列表(而不是reservationId对象)来获取客户
data class GuestWithReservations(
@Embedded val guest:Guest,
@Relation(
parentColumn = "id",
entityColumn = "guestId",
entity = ReservationGuest::class,
projection = "reservationId"
) val reservationIdList: List<Long>
)由于可以获得guestId和reservationId，因此可以使用它们查询Reservation和Guest实体。
如果我找到一种简单的方法来获取预订和访客对象列表，而不是它们的ids，我会更新我的答案。
Stack Overflow用户
发布于 2019-10-17 12:09:24
通过在房间中介绍Junction，您可以轻松处理多对多关系。
正如@Devrim所说，您可以使用一个额外的实体(ReservationGuest)来保持客户和预订之间的关系(也称为关联表、连接表或连接表)。
@Entity
data class Guest(
@PrimaryKey
val gId: Long,
val name: String,
val email: String
)
@Entity
data class Reservation(
@PrimaryKey
val rId: Long,
val table: String
)
@Entity(
primaryKeys = ["reservationId", "guestId"]
)
data class ReservationGuest(
val reservationId: Long,
val guestId: Long
)现在，您可以使用此模型与客人进行预订：
data class ReservationWithGuests (
@Embedded
val reservation: Reservation,
@Relation(
parentColumn = "rId",
entity = Guest::class,
entityColumn = "gId",
associateBy = Junction(
value = ReservationGuest::class,
parentColumn = "reservationId",
entityColumn = "guestId"
)
)
val guests: List<Guest>
)您也可以通过客人的预订列表获得他们的预订。
data class GuestWithReservations (
@Embedded
val guest: Guest,
@Relation(
parentColumn = "gId",
entity = Reservation::class,
entityColumn = "rId",
associateBy = Junction(
value = ReservationGuest::class,
parentColumn = "guestId",
entityColumn = "reservationId"
)
)
val reservations: List<Reservation>
)现在您可以在数据库中查询结果，如下所示：
@Dao
interface GuestReservationDao {
@Query("SELECT * FROM Reservation")
fun getReservationWithGuests(): LiveData<List<ReservationWithGuests>>
@Query("SELECT * FROM Guest")
fun getGuestWithReservations(): LiveData<List<GuestWithReservations>>
}Stack Overflow用户
发布于 2018-05-27 17:31:32
实际上还有一种获取Guest列表的可能性，而不仅仅是@Devrim答案中的id。
首先定义表示Guest和Reservation之间连接的类。
@Entity(primaryKeys = ["reservationId", "guestId"],
foreignKeys = [
ForeignKey(entity = Reservation::class,
parentColumns = ["id"],
childColumns = ["reservationId"]),
ForeignKey(entity = Guest::class,
parentColumns = ["id"],
childColumns = ["guestId"])
])
data class ReservationGuestJoin(
val reservationId: Long,
val guestId: Long
)每次要插入新的Reservation时，都必须插入ReservationGuestJoin对象以满足外键约束。现在，如果你想得到Guest列表，你可以使用SQL query的强大功能：
@Dao
interface ReservationGuestJoinDao {
@SuppressWarnings(RoomWarnings.CURSOR_MISMATCH)
@Query("""
SELECT * FROM guest INNER JOIN reservationGuestJoin ON
guest.id = reservationGuestJoin.guestId WHERE
reservationGuestJoin.reservationId = :reservationId
""")
fun getGuestsWithReservationId(reservationId: Long): List<Guest>
}要查看更多详细信息，请访问this blog。
https://stackoverflow.com/questions/44361824
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