在房间 android 中插入关系

Question

我对如何在房间中插入关系有疑问。所以，我有 Product 和 ProductsList 实体。 ProductsList 可能有很多产品，但产品不应该知道它们包含在何处的列表。

@Entity(tableName = "products")
data class Product(
    @PrimaryKey(autoGenerate = true)
    val productId: Long,

    val productName: String
)

@Entity
data class ProductList(
    @PrimaryKey(autoGenerate = true)
    val productListId: Long,
    val listName: String
)

我创建了 ProductsListWithProducts class:

data class ProductListWithProducts(
    @Embedded
    val productList: ProductList,

    @Relation(
        parentColumn = "productListId",
        entityColumn = "productId",
        entity = Product::class
    )
    val productsId: List<Product>
)

但我不明白如何在数据库中插入数据。例如，我已经在它的 table 中添加了 Products，然后它想创建新的 ProductList。我检查了其他答案，发现它只是使用 Dao 来插入它，比如：

@Dao
abstract class ProductListDao {

    @Transaction
    fun insert(productList: ProductList, products: List<Product>) {
        insert(productList)
        for (product in products) {
            insert(product)
        }
    }

但我不知道如何在这个 table 之间添加关系，因为我不想在产品实体中使用外键（因为在这种情况下我需要创建多对多关系).我考虑过其他实体

@Entity(primaryKeys = ["productId", "productListId"])
data class ProductListProducts(
    val productId: Long,
    val productListId: Long
)

但它也用于定义多对多关系。

我可以添加这个关系而不创建多对多关系，只创建一对多吗？

Answer 1

有 ProductListProducts 但是您可能希望考虑使用 :-

@Entity(
    primaryKeys = ["productId", "productListId"]
    ,indices = [
        Index(value = ["productListId"]) /* Index else Room warns */
    ]
    /* Foreign Keys are optional BUT enforce referential integrity */
    , foreignKeys = [
        ForeignKey(
            entity = Product::class,
            parentColumns = ["productId"],
            childColumns = ["productId"],
            onDelete = ForeignKey.CASCADE,
            onUpdate = ForeignKey.CASCADE
        ),
        ForeignKey(
            entity = ProductList::class,
            parentColumns = ["productListId"],
            childColumns = ["productListId"],
            onDelete = ForeignKey.CASCADE,
            onUpdate = ForeignKey.CASCADE
        )
    ]
    )
data class ProductListProducts(
    val productId: Long,
    val productListId: Long
)

您不妨参考https://sqlite.org/foreignkeys.html

这是一个关联 table（参考table、映射table和许多其他术语）。因此，您使用的关系利用 association 作为 ProductList 和 Product 之间的 Junction。因此，您的 ProductListWithProducts POJO 变为：-

data class ProductListWithProducts (
    @Embedded
    val productList: ProductList,
    @Relation(
        entity = Product::class,
        parentColumn = "productListId",
        entityColumn = "productId",
        associateBy = Junction(
            ProductListProducts::class,
            parentColumn = "productListId",
            entityColumn = "productId"
        )
    )
    val product: List<Product>
)

演示使用上述 classes（以及您的 classes，其中 id 已更改为 Long=0）

有道class喜欢:-

@Dao
abstract class AllDao {
    @Insert
    abstract fun insert(product: Product): Long
    @Insert
    abstract fun insert(productList: ProductList): Long
    @Insert
    abstract fun insert(productListProducts: ProductListProducts): Long
    @Transaction
    @Query("SELECT * FROM ProductList")
    abstract fun getProductListWithProducts(): List<ProductListWithProducts>
}

然后是下面的（运行在主线程上 brevity/convenience）：-

    db = TheDatabase.getInstance(this)
    dao = db.getAllDao()

    var p1 = dao.insert(Product( productName = "Product1"))
    var p2 = dao.insert(Product(productName = "Product2"))
    var pl1 = dao.insert(ProductList(listName = "List1"))
    var pl2 = dao.insert(ProductList(listName = "List2"))
    dao.insert(ProductListProducts(p1,pl1))
    dao.insert(ProductListProducts(p1,pl2))
    dao.insert(ProductListProducts(p2,pl1))

    dao.insert(ProductListProducts(dao.insert(Product(productName = "Product3")),dao.insert(
        ProductList(listName = "List3")))
    )
    for(plwp: ProductListWithProducts in dao.getProductListWithProducts()) {
        Log.d(TAG,"ProductList is ${plwp.productList.listName} ID is ${plwp.productList.productListId}")
        for(p: Product in plwp.product) {
            Log.d(TAG,"\t Product is ${p.productName} ID is ${p.productId}")
        }
    }

结果日志中包含：-

D/DBINFO: ProductList is List1 ID is 1
D/DBINFO:    Product is Product1 ID is 1
D/DBINFO:    Product is Product2 ID is 2
D/DBINFO: ProductList is List2 ID is 2
D/DBINFO:    Product is Product1 ID is 1
D/DBINFO: ProductList is List3 ID is 3
D/DBINFO:    Product is Product3 ID is 3

Can I add this relation without creating many-to-many relation, just one-to-many?

以上（即关联 table）将处理 1 到多，但如果您不想要额外的 table，则必须包含 [=69 的标识符=] 在 child 中。您可以在关联 table 中强制执行一对多，方法是使 1 的列唯一。

至于你说的Mass类型插入

For example I already added Products in its table and after it want to create new ProductList

那么你也许可以通过以下额外的@Dao 来解决问题:-

@Insert
abstract fun insert(productListList: List<ProductList>): LongArray
@Insert
abstract fun insertManyProductListProducts(productListProductsList: List<ProductListProducts>): LongArray



/* This can be used to get a specific product or products according to a pattern */
/* e.g. */
/*  if productPatterName is product1 then an exact match */
/* if prod% then all that start with prod */
/* if %prod all that end in prod */
/* if %prod% then all that have prod anywhere */
@Query("SELECT productId FROM products WHERE productName LIKE :productNamePattern")
abstract fun getProductIdByName(productNamePattern: String): LongArray

然后有代码如:-

    /* Adding many new ProductLists to existing Products */
    /* 1 add the new ProductLists */
    /*      Noting that the insert returns an array of the productListId's inserted */
    val insertedProductLists = dao.insert(
        listOf(
            ProductList(listName = "ListX1"),
            ProductList(listName = "ListX2")
        )
    )
    /* 2. Determine the Product(s) that will be related to the new list of ProductLists */
    val productIdList = dao.getProductIdByName("Product%") /* All products */
    /* 3. Prepare the List of ProductListProducts for mass insertion */
    val plplist: ArrayList<ProductListProducts> = ArrayList()
    for(pid: Long in productIdList) {
        for(plid: Long in insertedProductLists) {
            plplist.add(ProductListProducts(pid,plid))
        }
    }
    /* 4. add the relationships */
    dao.insertManyProductListProducts(plplist)

如果您想要 1 个现有产品，例如product1 那么你就用dao.getProductIdByName("Product1")，当然你也可以increase/reduceArray中的productLists来适应。

这将导致：-

D/DBINFO: ProductList is List1 ID is 1
D/DBINFO:    Product is Product1 ID is 1
D/DBINFO:    Product is Product2 ID is 2
D/DBINFO: ProductList is List2 ID is 2
D/DBINFO:    Product is Product1 ID is 1
D/DBINFO: ProductList is List3 ID is 3
D/DBINFO:    Product is Product3 ID is 3
D/DBINFO: ProductList is ListX1 ID is 4
D/DBINFO:    Product is Product1 ID is 1
D/DBINFO:    Product is Product2 ID is 2
D/DBINFO:    Product is Product3 ID is 3
D/DBINFO: ProductList is ListX2 ID is 5
D/DBINFO:    Product is Product1 ID is 1
D/DBINFO:    Product is Product2 ID is 2
D/DBINFO:    Product is Product3 ID is 3

在房间 android 中插入关系

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android-room