How can I represent a "many to many" relation with Android Room when column names are same?
Solution 1:
I had a similar issue. Here is my solution.
You can use an extra entity (ReservationGuest
) which keeps the relation between Guest
and Reservation
.
@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
)
You can get reservations with their list of guestId
s. (Not the guest objects)
data class ReservationWithGuests(
@Embedded val reservation:Reservation,
@Relation(
parentColumn = "id",
entityColumn = "reservationId",
entity = ReservationGuest::class,
projection = "guestId"
) val guestIdList: List<Long>
)
You can also get guests with their list of reservationId
s. (Not the reservation objects)
data class GuestWithReservations(
@Embedded val guest:Guest,
@Relation(
parentColumn = "id",
entityColumn = "guestId",
entity = ReservationGuest::class,
projection = "reservationId"
) val reservationIdList: List<Long>
)
Since you can get the guestId
s and reservationId
s, you can query Reservation
and Guest
entities with those.
I'll update my answer if I find an easy way to fetch Reservation and Guest object list instead of their ids.
Similar answer
Solution 2:
With the introduction to Junction in room you can handle many-to-many relationship with ease.
As @Devrim stated you can use an extra entity (ReservationGuest) which keeps the relation between Guest and Reservation(also know as associative table or junction table or join table).
@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
)
Now you can get reservation with guests using this model:
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>
)
You can also get guest with their list of reservations as.
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>
)
Now you can query database for the result as:
@Dao
interface GuestReservationDao {
@Query("SELECT * FROM Reservation")
fun getReservationWithGuests(): LiveData<List<ReservationWithGuests>>
@Query("SELECT * FROM Guest")
fun getGuestWithReservations(): LiveData<List<GuestWithReservations>>
}
Solution 3:
Actually there is one more possibility to get Guest
list, not only id's like in @Devrim answer.
First define class which will represent the connection between Guest
and Reservation
.
@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
)
Each time you will be inserting new Reservation
, you will have to insert ReservationGuestJoin
object in order to fulfill foreign key constraint.
And now if you want to get Guest
list you can use power of SQL query:
@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>
}
To see more details visit this blog.
Solution 4:
Here is a way to query a full object model through an M:N junction table in a single query. The subqueries are probably not the most efficient way to do this, but it does work until they get @Relation
to properly walk through ForeignKey
. I hand-jammed the Guest/Reservation framework into my working code so there may be typos.
Entity (This has been covered)
@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
)
Dao (Note we pull in the M:N via a subquery and reduce the extra Reservation
rows with a GROUP_CONCAT
@Query("SELECT *, " +
"(SELECT GROUP_CONCAT(table) " +
"FROM ReservationGuest " +
"JOIN Reservation " +
"ON Reservation.id = ReservationGuest.reservationId " +
"WHERE ReservationGuest.guestId = Guest.id) AS tables, " +
"FROM guest")
abstract LiveData<List<GuestResult>> getGuests();
GuestResult (This handles the mapping of the query result, note we convert the concatenated string back to a list with @TypeConverter
)
@TypeConverters({ReservationResult.class})
public class GuestResult extends Guest {
public List<String> tables;
@TypeConverter
public List<String> fromGroupConcat(String reservations) {
return Arrays.asList(reservations.split(","));
}
}