我正在尝试在 3 个彼此按顺序一对多的表之间创建关系。所以表 1 与表 2 是多对一的，而表 2 与表 3 是多对一的。
我要做的是创建一个关系，这样我就可以拥有 Table3 以及相关的 Table2 实体列表以及与 Table2 相关的另一个 Table1 实体列表。我一直在查看此处的文档和问题，但我尝试过的关系要么给了我例外，要么缺少实体，要么提供了与我查询的内容无关的实体。
如果按照我尝试的方式无法做到这一点，您能否指导我朝着正确的方向前进？我目前拥有的是一个事务，它查询第一个关系（Table3-Table2），然后根据第一个实体（Table2-Table1）查询第二个关系，并返回所有这些关系的对象。
如果已经有问题/博客文章回答了这个问题，请告诉我，因为我找不到它。
【问题讨论】：
标签： android android-room one-to-many
您必须以分层方式（自下而上）处理此问题
您有一个用于 Table2 的 POJO (@Embedded)，其中包含一个 Table3 子级列表 (@Relation)
然后，您将拥有 Table1 (@Embedded) 的 POJO，其中包含 Table2 POJO 子项 (@Relation) 的列表。
例如:-
表 1
@Entity(tableName = "table1")
data class Table1(
@PrimaryKey
@ColumnInfo(name = "table1_id")
val id: Long? = null,
@ColumnInfo(name = "table1_name")
val name: String
)
表2
@Entity(
tableName = "table2",
foreignKeys = [
ForeignKey(
entity = Table1::class,
parentColumns = ["table1_id"],
childColumns = ["map_to_parent_in_table1"],
onDelete = CASCADE,
onUpdate = CASCADE
)
]
)
data class Table2(
@PrimaryKey
@ColumnInfo(name = "table2_id")
val id: Long? = null,
@ColumnInfo(name = "map_to_parent_in_table1", index = true)
val map_to_parent: Long,
@ColumnInfo(name = "table2_name")
val name: String
)
Table3
@Entity(
tableName = "table3",
foreignKeys = [
ForeignKey(
entity = Table2::class,
parentColumns = ["table2_id"],
childColumns = ["map_to_parent_in_table2"],
onDelete = ForeignKey.CASCADE,
onUpdate = ForeignKey.CASCADE
)
]
)
data class Table3(
@PrimaryKey
@ColumnInfo(name = "table3_id")
val id: Long? = null,
@ColumnInfo(name = "map_to_parent_in_table2", index = true)
val map_to_parent: Long,
@ColumnInfo(name = "table3_name")
val name: String
)
foreignKeys 是可选的，但可以强制执行并帮助维护参照完整性。在外键中，onDelete 和 onUpdate 是可选的所以从最低的（有孩子的最小的兄弟姐妹，即 Table2）开始，然后是 POJO
Table2WithTable3Children
data class Table2WithTable3Children(
@Embedded
val table2: Table2,
@Relation(
entity = Table3::class,
parentColumn = "table2_id",
entityColumn = "map_to_parent_in_table2"
)
val children: List<Table3>
)
然后是 POJO
Table1WithTable2ChildrenIncludingTable3Children
data class Table1WithTable2ChildrenIncludingTable3Children(
@Embedded
val table1: Table1,
@Relation(
entity = Table2::class,
parentColumn = "table1_id",
entityColumn = "map_to_parent_in_table1"
)
val children: List<Table2WithTable3Children>
)
所以如果你得到一个 Table1WithTable2ChildrenIncludingTable3Children 那么它包含一个 Table2WithTable3Children 列表。
你可以有这样的查询：-
@Query("SELECT * FROM table1" )
@Transaction
abstract fun getAllTable1sWithTable2ChildrenAndTable3Children(): List<Table1WithTable2ChildrenIncludingTable3Children>
这可以像这样使用：-
dao.getAllTable1sWithTable2ChildrenAndTable3Children()
for(t1: Table1WithTable2ChildrenIncludingTable3Children in dao.getAllTable1sWithTable2ChildrenAndTable3Children()) {
Log.d(TAG,"Table1 is ${t1.table1.name} ID is ${t1.table1.id} it has ${t1.children.size} children in Table2 they are:-")
for(t2: Table2WithTable3Children in t1.children) {
Log.d(TAG,"\tTable2 is ${t2.table2.name} ID is ${t2.table2.id} it has ${t2.children} in Table3 they are :-")
for(t3: Table3 in t2.children) {
Log.d(TAG,"\t\tTable3 is ${t3.name} ID is ${t3.id}")
}
}
}
工作示例/演示
使用@Dao 注释类AllDao 扩展上述内容（包括上面的查询）：-
@Dao
abstract class AllDao {
@Insert
abstract fun insert(table1: Table1): Long
@Insert
abstract fun insert(table2: Table2): Long
@Insert
abstract fun insert(table3: Table3): Long
@Query("SELECT * FROM table1" )
@Transaction
abstract fun getAllTable1sWithTable2ChildrenAndTable3Children(): List<Table1WithTable2ChildrenIncludingTable3Children>
}
和一个基本的@Database 注释类ThisDatabase :-
@Database(entities = [Table1::class,Table2::class,Table3::class], exportSchema = false, version = 1)
abstract class ThisDatabase: RoomDatabase() {
abstract fun getAllDao(): AllDao
companion object {
private var instance: ThisDatabase? = null
fun getInstance(context: Context): ThisDatabase {
if (instance == null) {
instance = Room.databaseBuilder(context,ThisDatabase::class.java,"this_database.db")
.allowMainThreadQueries()
.build()
}
return instance as ThisDatabase
}
}
}
并测试一下 MainActivity :-
class MainActivity : AppCompatActivity() {
lateinit var db: ThisDatabase
lateinit var dao: AllDao
val TAG = "DBINFO"
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
setContentView(R.layout.activity_main)
db = ThisDatabase.getInstance(this)
dao = db.getAllDao()
val t1_1 = dao.insert(Table1( name = "First Table1"))
val t1_2 = dao.insert(Table1(name = "Second Table1"))
val t1_3 = dao.insert(Table1(name = "Third Table1"))
val t2_1 = dao.insert(Table2(name = "First Table2 with First Table1 as Parent", map_to_parent = t1_1))
val t2_2 = dao.insert(Table2(name = "Second Table2 with First Table1 as Parent", map_to_parent = t1_1))
val t2_3 = dao.insert(Table2(name = "Third Table2 with Second Table1 as Parent", map_to_parent = t1_2))
dao.insert(Table3(name = "T31", map_to_parent = t2_1))
dao.insert(Table3(name = "T32", map_to_parent = t2_2))
dao.insert(Table3(name = "T33", map_to_parent = t2_3))
dao.insert(Table3(name = "T34", map_to_parent = t2_1))
dao.insert(Table3(name = "T35", map_to_parent = t2_1))
dao.getAllTable1sWithTable2ChildrenAndTable3Children()
for(t1: Table1WithTable2ChildrenIncludingTable3Children in dao.getAllTable1sWithTable2ChildrenAndTable3Children()) {
Log.d(TAG,"Table1 is ${t1.table1.name} ID is ${t1.table1.id} it has ${t1.children.size} children in Table2 they are:-")
for(t2: Table2WithTable3Children in t1.children) {
Log.d(TAG,"\tTable2 is ${t2.table2.name} ID is ${t2.table2.id} it has ${t2.children.size} in Table3 they are :-")
for(t3: Table3 in t2.children) {
Log.d(TAG,"\t\tTable3 is ${t3.name} ID is ${t3.id}")
}
}
}
}
}
那么第一次运行时写入日志的Result是：-
D/DBINFO: Table1 is First Table1 ID is 1 it has 2 children in Table2 they are:-
D/DBINFO: Table2 is First Table2 with First Table1 as Parent ID is 1 it has 3 in Table3 they are :-
D/DBINFO: Table3 is T31 ID is 1
D/DBINFO: Table3 is T34 ID is 4
D/DBINFO: Table3 is T35 ID is 5
D/DBINFO: Table2 is Second Table2 with First Table1 as Parent ID is 2 it has 1 in Table3 they are :-
D/DBINFO: Table3 is T32 ID is 2
D/DBINFO: Table1 is Second Table1 ID is 2 it has 1 children in Table2 they are:-
D/DBINFO: Table2 is Third Table2 with Second Table1 as Parent ID is 3 it has 1 in Table3 they are :-
D/DBINFO: Table3 is T33 ID is 3
D/DBINFO: Table1 is Third Table1 ID is 3 it has 0 children in Table2 they are:-
【讨论】：