Room 2.5一对一多对一嵌套关系

问题描述

房间

我有这样的关系 拼图->有一个->对话->有很多->对话行

我遵循了 https://developer.android.com/training/data-storage/room/relationships#kotlin 但仍然无法正常工作

这些似乎与错误有关：

错误：在com.example.puzzleherexamenandroid.data.room.databaseModels.DialogueWithLinesDatabase中找不到子实体列puzzleDialogueId。

尝试了以下构造函数，但它们不匹配： PuzzleWithDialogueDatabase（com.example.puzzleherexamenandroid.data.room.databaseModels.PuzzleDatabase，com.example.puzzleherexamenandroid.data.room.databaseModels.DialogueWithLinesDatabase）-> [param：puzzle->匹配字段：puzzle，param：dialogueWithLines->匹配字段：unmatched] C：\ Users \ Jasper \ StudioProjects \ PuzzleHerexamenAndroid \ app \ build \ tmp \ kapt3 \ stubs \ debug \ com \ example \ puzzleherexamenandroid \ data \ room \ databaseModels \ PuzzleWithDialogueDatabase.java：9：错误：找不到以下设置器字段。

PuzzleDatabaseDao.java:12：错误：参数的类型必须是带有@Entity或其集合/数组注释的类。 java.util.List 拼图）；

这些是我的实体

@Entity(tableName = "puzzle_table")
@Parcelize
data class PuzzleDatabase (
     @PrimaryKey
     val puzzleId: Int,val title: String,val prompt: String,val answer: String

): Parcelable 

 @Entity(tableName = "dialogue_table")
@Parcelize
data class DialogueDatabase (
    @PrimaryKey
    val dialogueId: Int,val char1avatar: String,val char2avatar: String,val puzzleDialogueId : Int
): Parcelable

@Entity(tableName = "dialogueLine_table")
@Parcelize
data class DialogueLineDatabase (
    @PrimaryKey
    val dialogueLineId: Int,val line: String,val speaking: Int,val dialogueForeignkeyId: Int
    ): Parcelable

这些是关系的类

data class PuzzleWithDialogueDatabase(
    @Embedded val puzzle : PuzzleDatabase,@Relation(
        entity = DialogueWithLinesDatabase::class,parentColumn = "puzzleId",entityColumn = "puzzleDialogueId"
    )
    val dialogueWithLines: DialogueWithLinesDatabase
)

data class DialogueWithLinesDatabase(
    @Embedded val dialogue:DialogueDatabase,@Relation(
        parentColumn = "dialogueId",entityColumn = "dialogueForeignkeyId"
    )
    val dialogueLines: List<DialogueLineDatabase>
)

这是我的事

@Dao
interface PuzzleDatabaseDao {
   @Transaction
   @Insert(onConflict = OnConflictStrategy.IGNORE)
   fun insertAll(puzzle: List<PuzzleWithDialogueDatabase>)

   @Transaction
   @Query("SELECT * from puzzle_table ORDER BY puzzleId DESC")
   fun getAllPuzzles(): LiveData<List<PuzzleWithDialogueDatabase>>
}

解决方法

我想您应该在DialogueWithLinesDatabase类的“版本”定义中用DialogueDatabase更改实体PuzzleWithDialogueDatabase：

data class PuzzleWithDialogueDatabase(
    @Embedded val puzzle : PuzzleDatabase,@Relation(
        entity = DialogueDatabase::class,// <- changed
        parentColumn = "puzzleId",entityColumn = "puzzleDialogueId"
    )
    val dialogueWithLines: DialogueWithLinesDatabase
)

,

@Relation中的实体属性应提及数据库实体，而不是关系类。试试：

data class PuzzleWithDialogueDatabase(
    @Embedded val puzzle : PuzzleDatabase,parentColumn = "puzzleId",entityColumn = "puzzleDialogueId"
    )
    val dialogueWithLines: DialogueWithLinesDatabase
)