如何在Android房间中拥有ID的独特字段？

Question

我想为ID实现唯一的AUDIO_ID。

这是我的实体课程

data class Members(
    @PrimaryKey(autoGenerate = true) @ColumnInfo(name = "id") val id: Int?,
    var title: String,
    var artist: String,
    var album: String,
    @ColumnInfo(name = "audio_id") val audioId: Int,
    val albumId: String
)

不起作用

@Entity(tableName = "playlist_members", indices = [Index(value = ["id","audio_id"], unique = true)])

:)

Answer 1

我做了一个测试示例。所以所有的工作。

我的实体

@Entity(tableName = "cards", indices = [Index(value = ["data"], unique = true)])
data class CardEntity(
   @PrimaryKey(autoGenerate = true)
   val id: Int = 0,
   val data: String
)

我的测试

@Test
fun testRun() = runBlocking {
    db.cardDAO().insert(CardEntity(data = "1"))
    var exception = false
    try {
        db.cardDAO().insert(CardEntity(data = "1"))
    } catch (e: SQLiteConstraintException) {
        exception=true
    }
    db.cardDAO().insert(CardEntity(data = "2"))
    Assert.assertTrue(exception)
    Assert.assertEquals(db.cardDAO().getAll().first().size, 2 )
}

，但您想获得ID唯一的AUDIO_ID。也许一样。

data class Members(
   @PrimaryKey(autoGenerate = true) @ColumnInfo(name = "audio_id") val audioId = 0,
   var title: String,
   var artist: String,
   var album: String,
   val albumId: String
)