要在Android Room中实现在一个对象中插入另一个对象的功能,你可以使用嵌套关系(Nested Relationship)来实现。下面是一个示例代码:
首先,假设你有两个实体类:CardSet(闪卡包)和Question(问题)。
@Entity(tableName = "card_sets")
data class CardSet(
@PrimaryKey(autoGenerate = true)
val id: Int = 0,
val name: String
)
@Entity(tableName = "questions", foreignKeys = [ForeignKey(entity = CardSet::class,
parentColumns = ["id"],
childColumns = ["cardSetId"],
onDelete = ForeignKey.CASCADE)])
data class Question(
@PrimaryKey(autoGenerate = true)
val id: Int = 0,
val cardSetId: Int,
val questionText: String
)
然后,在CardSet实体类中,你可以使用@Relation注解来指定CardSet和Question之间的关系。
data class CardSetWithQuestions(
@Embedded val cardSet: CardSet,
@Relation(
parentColumn = "id",
entityColumn = "cardSetId"
)
val questions: List
)
接下来,在你的Dao接口中,你可以使用@Query注解来编写查询语句,用于获取包含问题的闪卡包。
@Dao
interface CardSetDao {
@Transaction
@Query("SELECT * FROM card_sets")
fun getCardSetsWithQuestions(): List
}
最后,在你的数据库类中,你需要将CardSet和Question的实体类添加到entities参数中,并在数据库的版本更新时执行对应的迁移操作。
@Database(entities = [CardSet::class, Question::class], version = 1)
abstract class AppDatabase : RoomDatabase() {
abstract fun cardSetDao(): CardSetDao
}
这样,你就可以在一个对象中插入另一个对象了。例如,你可以通过以下方式将一个CardSet对象和包含的Question对象插入到数据库中:
val cardSet = CardSet(name = "My Card Set")
val questions = listOf(
Question(cardSetId = 1, questionText = "Question 1"),
Question(cardSetId = 1, questionText = "Question 2")
)
val cardSetWithQuestions = CardSetWithQuestions(cardSet = cardSet, questions = questions)
// 插入CardSet和关联的Questions
database.cardSetDao().insertCardSetWithQuestions(cardSetWithQuestions)
希望这个示例能帮助到你!