Android 删除一对多关系室 Kotlin 中的所有值

您应该使用ForeignKey来指定实体之间的关系.

正如文档所说:

外键允许您指定跨实体的约束，如 该SQLite将确保在您修改时关系有效 数据库.

要添加外键，请执行以下操作:

1. 父类

@Entity
data class User(
    @PrimaryKey
    val userId: Int,
    val name: String
)

2. children 班:

@Entity(
    foreignKeys = [
        ForeignKey(
            entity = User::class,
            parentColumns = arrayOf("userId"),
            childColumns = arrayOf("ownerUserId"),
            onDelete = ForeignKey.CASCADE
        )
    ]
)
data class Pet(
    @PrimaryKey
    val petId: Int,
    val name: String,
    @ColumnInfo(index = true)
    val ownerUserId: Int
)

3. 定义对象之间的关系:

data class UserWithPets (
    @Embedded
    val user: User,
    @Relation(
        parentColumn = "userId",
        entityColumn = "ownerUserId"
    )
    val pets: List<Pet>
)

4. 创建DAO:

@Dao
interface UserDao {
    @Insert
    suspend fun saveUser(user: User)

    @Insert
    suspend fun saveUsers(users: List<User>)

    @Insert
    suspend fun savePets(pets: List<Pet>)

    @Query("SELECT * FROM USER")
    suspend fun getUsers(): List<UserWithPets>

    @Query("DELETE FROM USER WHERE userId=:id")
    suspend fun deleteUser(id: Int)

    @Query("DELETE FROM USER")
    suspend fun deleteAllUsers()

    @Update
    suspend fun updatePet(pet: Pet)
}

Some example queries:个

// Add new user
 val user = User(userId = 1, name = "User1")
 userDao.saveUser(user)

 var userWithPets = userDao.getUsers()
// result -> UserWithPets(user=User(userId=1, name=User1), pets=[])

// Add new pet
 val pet = Pet(petId = 1, ownerUserId = 1, name = "Pet Name")
 userDao.savePets(listOf(pet))

// Fetch users again
//UserWithPets(user=User(userId=1, name=User1), pets=[Pet(petId=1, name=Pet Name, ownerUserId=1)])

 
// Update pet
userDao.updatePet(pet.copy(name = "New Name"))

// Fetch users again
//UserWithPets(user=User(userId=1, name=User1), pets=[Pet(petId=1, name=New Name, ownerUserId=1)])

这样，无论何时删除User，它的所有Pets都将自动删除.