如何将数据插入多对多关系表并检索最后一个 ID 在该方法接受两个参数的情况下？

Question

当输入只有一个参数时，Room return 插入的 id

  @Insert(onConflict = OnConflictStrategy.REPLACE)
    suspend fun insertWithDrinks(item: DrinkFavourite): Long

当方法看起来像这样时，我如何检索 ID

 @Insert(onConflict = OnConflictStrategy.REPLACE)
    suspend fun insertWithDrinks(item: DrinkFavourite, List<Drink>)

或者更确切地说，你如何处理多对多关系插入？我有这样的关系和交叉引用 table


data class DrinkFavouritesWithDrinks(
    @Embedded
    val item: DrinkFavourite,
    @Relation(
        parentColumn = "item_id",
        entityColumn = "drink_id",
        associateBy = Junction(DrinkFavouriteAndDrinksCrossRef::class)
    )
    val drinks: List<Drink> = emptyList()
)

我的交叉引用

@Entity(
    primaryKeys = ["item_id", "drink_id"]
)
data class DrinkFavouriteAndDrinksCrossRef(
    @ColumnInfo(name = "item_id")
    val itemId: Long,
    @ColumnInfo(name = "drink_id")
    val drinkId: Int
)

Answer 1

How can i retrieve the ID when method looks like this

当您插入多行时，您会在 Array<Long>

中获得结果 ID

但是，对于基本交叉 reference/mapping table，您很少会使用 rowid 列（即返回值）。

rowid 即使该列是隐式定义的，也会返回。也就是说，除了很少使用的 WITHOUT ROWID tables 之外，所有 tables 都有一个名为 rowid 的隐藏列。
- 当您使用INTEGER PRIMARY KEY定义列时，该列是rowid列的别名。在房间 @PrimaryKey 中，对于 Long 或 Int，有或没有 autoGenerate = true/false（没有 = false）等于 INTEGER PRIMARY KEY 或者如果 autoGenerate = true 等于 INTEGER PRIMARY KEY AUTOINCREMENT.

How do you handle many-to-many relationship Insert?

这是一个基于可用代码并创建了其他代码的示例。

所以除了你的 DrinkFavouriteAndDrinksCrossRef 实体。

DrinkFavourite 和 Item 的实体是

@Entity
 data class DrinkFavourite(
    @PrimaryKey
    val drink_id: Int? = null,
    val drink_name: String = ""
)
@Entity

数据class饮料( @PrimaryKey var itemId : 长?, 变种项目名称：字符串 )

The Dao AllDao 是:-

@Dao
interface AllDao {

    @Insert
    fun insertDrinkFavourite(drinkFavourite: DrinkFavourite) :Long

    @Insert
    fun insertManyDrinkFavourite(drinkFavourites: List<DrinkFavourite>) :Array<Long>

    @Insert
    fun insertItem(drink: Drink) :Long

    @Insert
    fun insertManyItems(drinks: List<Drink>) :Array<Long>

    @Insert(onConflict = OnConflictStrategy.IGNORE)
    fun insertDrinkFavouriteAndDrinksCrossRef(drinkFavouriteAndDrinksCrossRef: DrinkFavouriteAndDrinksCrossRef) :Long

    @Insert(onConflict = OnConflictStrategy.IGNORE)
    fun insertManyDrinkFavouriteAndDrinksCrossRef(drinkFavouriteAndDrinksCrossRefs: List<DrinkFavouriteAndDrinksCrossRef>) :Array<Long>
}

这适合为 3 个实体中的每一个插入一个 Entity/row 和多个 Entity/row。

下面利用道中定义的3多插入插入3Drink、4DrinkFavourite再插入12DrinkFavouriteAndDrinksCrossRef（3Drink * 4 DrinkFavorite) 排列，即所有可能的排列，可以根据插入的 id 进行交叉引用。

class MainActivity : AppCompatActivity() {

    lateinit var database: Database

    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
        setContentView(R.layout.activity_main)

        database = Room.databaseBuilder(this, Database::class.java, "drinksdb")
            .allowMainThreadQueries()
            .build()
        //val firsItemtid = database.allDao().insertItem( Item(0,"X"))
        val manyItemIds = database.allDao().insertManyItems(
            listOf(
                Drink(null,"A"),
                Drink(null,"B"),
                Drink(null,"C")
            )
        )
        val manyDrinkFavouriteIds = database.allDao().insertManyDrinkFavourite(
            listOf(
                DrinkFavourite(null,"DrinkA"),
                DrinkFavourite(null,"DrinkB"),
                DrinkFavourite(null,"DrinkC"),
                DrinkFavourite(null,"DrinkD")
            )
        )

        var xrefcombos = ArrayList<DrinkFavouriteAndDrinksCrossRef>()
        for(itemId: Long in manyItemIds) {
            for (dfId: Long in manyDrinkFavouriteIds) {
                xrefcombos.add( DrinkFavouriteAndDrinksCrossRef(itemId,dfId.toInt()))
            }
        }
        val drinkFavouriteAndDrinksCrossRefIdList = database.allDao().insertManyDrinkFavouriteAndDrinksCrossRef(xrefcombos)
        for (dfadcrId in drinkFavouriteAndDrinksCrossRefIdList) {
            Log.d("DRKFAVDRNKXREF","Id = " + dfadcrId)
        }
    }
}

输出为：-

2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 1
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 2
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 3
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 4
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 5
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 6
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 7
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 8
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 9
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 10
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 11
2020-01-17 07:46:40.621 D/DRKFAVDRNKXREF: Id = 12

即添加了 12 个排列。

除了根据插入的 Entity/Object 进行插入之外，我不相信您可以有效地使用 @Insert 进行其他操作。要根据其他值进行更复杂的插入，您可以使用 @Query 和 suitable SQL.

额外

关于 :-

data class DrinkFavouritesWithDrinks(
    @Embedded
    val item: DrinkFavourite,
    @Relation(
        parentColumn = "item_id",
        entityColumn = "drink_id",
        associateBy = Junction(DrinkFavouriteAndDrinksCrossRef::class)
    )

我认为这应该是:-

data class DrinkFavouriteWithDrinks (

    @Embedded
    val drinkFavourite: DrinkFavourite,
    @Relation(
        entity =  Drink::class,
        entityColumn = "itemId",
        parentColumn = "drink_id",
        associateBy = Junction(
            DrinkFavouriteAndDrinksCrossRef::class,entityColumn = "item_id",parentColumn = "drink_id"))
    val drinks: List<Drink> = emptyList()
    )

您可以获得每个 DrinkFavourite 的所有饮料，因此单数而不是复数是一个更好的名称。
关联需要知道要关联的列。

你可以有一个 Dao 方法，例如：-

@Query("SELECT * FROM drinkfavourite")
fun getAllDrinkFavouritesWithDrinks() :List<DrinkFavouriteWithDrinks>

将以下内容添加到上面的 MainActivity 中：-

    val drinkFavouriteWithDrinksList = database.allDao().getAllDrinkFavouritesWithDrinks()
    var sb = StringBuilder()
    for (dfwd: DrinkFavouriteWithDrinks in drinkFavouriteWithDrinksList) {
        sb.clear().append("DrinkFavourite = ").append(dfwd.drinkFavourite)
        for (d: Drink in dfwd.drinks) {
            sb.append("\n\tDrink = ").append(d.itemName)
        }
        Log.d("DRINKFAVINFO", sb.toString())
    }

结果：-

2020-01-17 11:39:52.762 D/DRINKFAVINFO: DrinkFavourite = DrinkFavourite(drink_id=1, drinkFavouriteName=DrinkA)
      Drink = A
      Drink = B
      Drink = C
2020-01-17 11:39:52.763 D/DRINKFAVINFO: DrinkFavourite = DrinkFavourite(drink_id=2, drinkFavouriteName=DrinkB)
      Drink = A
      Drink = B
      Drink = C
2020-01-17 11:39:52.763 D/DRINKFAVINFO: DrinkFavourite = DrinkFavourite(drink_id=3, drinkFavouriteName=DrinkC)
      Drink = A
      Drink = B
      Drink = C
2020-01-17 11:39:52.763 D/DRINKFAVINFO: DrinkFavourite = DrinkFavourite(drink_id=4, drinkFavouriteName=DrinkD)
      Drink = A
      Drink = B
      Drink = C

不是最好的例子，因为每个 DrinkFavourite 都有这三种饮料

如何将数据插入多对多关系表并检索最后一个 ID 在该方法接受两个参数的情况下？

How to Insert data to Many to many relationship tables and retrieving last id In cases where the method accepts two parameters?

android

dao

many-to-many

android-room

额外