java - 如何使用带有FK的表在查询室中获取完整的实体对象

Question

假设我有这些表:
表-用户

Stores the users

@Entity(
    tableName = "USER"
)
data class User(
    @PrimaryKey(autoGenerate = true)
    @ColumnInfo(name = "user_id")
    val id: Int,

    @ColumnInfo(name = "user_name")
    val name: String
)

表-项目

Store the items it's like a product

@Entity(
    tableName = "ITEM"
)
data class Item(
    @PrimaryKey(autoGenerate = true)
    @ColumnInfo(name = "item_id")
    val id: Int,

    @ColumnInfo(name = "item_name")
    val name: String,

    @ColumnInfo(name = "item_description")
    val description: String
)

表-特殊

Store an speciality for product 1 Special needs a Product to exist

@Entity(
    tableName = "SPECIAL",
    foreignKeys = [ForeignKey(
        entity = Item::class,
        parentColumns = ["item_id"],
        childColumns = ["special_item_id"]
    )]
)
data class Special(
    @PrimaryKey(autoGenerate = true)
    @ColumnInfo(name = "special_id")
    val id: Int,

    @ColumnInfo(name = "special_item_id")
    val coupon_product_id: Int,

    @ColumnInfo(name = "special_name")
    val name: String,

    @ColumnInfo(name = "special_description")
    val description: String

)

表-最喜欢的

Stores the favourite Specials from an user

@Entity(
    tableName = "TB_FAVOURITE",
    foreignKeys = [ForeignKey(
        entity = User::class,
        parentColumns = ["user_id"],
        childColumns = ["favourite_user_id"]
    ), ForeignKey(
        entity = Special::class,
        parentColumns = ["special_id"],
        childColumns = ["favourite_special_id"]
    )]
)
data class Favourite(
    @PrimaryKey
    @ColumnInfo(name = "favourite_user_id")
    val id: Int,

    @ColumnInfo(name = "favourite_special_id")
    val specialId: Int

)

我的问题是，如何查询以选择所有的Specials，然后创建类似于存储用户是否喜欢的类的类。目前，这是一个用户应用程序，是一个演示应用程序。因此，用户将始终相同，因此我可以对findById进行硬编码并发送用户ID。
目标
是获取查询结果的列表，其中包含:

所有特价商品

在SomeClass中应在

内包含Special的项目

一个标志，以了解它是否是该驱动程序的最爱

问题是我希望能够将 session 室数据库的结果映射到所需的对象，因此我想查询比映射器更重要，我知道该怎么做。
请注意，我使用的是assets/database/mydb.db文件来启动数据库，我不知道这是否重要。
查询将如何做到这一点？ db的结构是否有任何改进以使其更容易？

Answer 1

您可以这样创建类:

class SpecialWithFavourite {
    @Embedded
    var special: Special? = null
    @Relation(parentColumn = "special_item_id", entityColumn = "item_id")
    var item: Item? = null
    @Relation(parentColumn = "special_id", entityColumn = "favourite_special_id")
    var favourite: Favourite? = null
}

然后使用dao查询将其提取如下:

@Dao
interface DaoSpecialWithFavourite {
    @Transaction
    @Query("SELECT SPECIAL.* FROM SPECIAL INNER JOIN TB_FAVOURITE WHERE TB_FAVOURITE.favourite_user_id = :userId")
    fun getLD(userId: Long): List<SpecialWithFavourite>
}

另一种选择是使用View，特别是如果您只需要几个字段时

@DatabaseView("SELECT SPECIAL.*, ITEM.item_name FROM SPECIAL INNER JOIN ITEM ON ITEM.item_id = SPECIAL.special_item_id")
class FavSpecial {
    var name: String? = null
    @Embedded
    var special: Special? = null
}

根据评论中的讨论，新Favourite

data class Favourite(
    @PrimaryKey(autoGenerate = true)
    val id: Int,
    @ColumnInfo(name = "favourite_user_id")
    val userid: Int,
    @ColumnInfo(name = "favourite_special_id")
    val specialId: Int
)

这样，您仍然可以保留约束，并且用户可以拥有多个喜欢的特价商品