Question

我想在我的Room DB中存储字符串（uid）。我希望这些uid能够做到：

已查询（请检查我的字符串列表中是否存在uid）
添加到（向字符串列表添加uid）

最有效的方法是什么？

这里是给您一些背景知识的骨架，尽管我对Room的理解非常基础，所以会出现错误：

@Entity(tableName = "user_data")
data class UserData(
    @PrimaryKey(autoGenerate = true) val uid: Int,
    @ColumnInfo(name = "matched_users") var matchedUsers: Set<String>
)


@Dao
interface UserDataDao {
    // Check to see if the uid is in matchedUsers Set/List
    @Query("SELECT * FROM user_data WHERE :matchId IN matched_users")
    fun matchedBefore(matchId: String): Boolean

    // Add uid to Set/List
    @Insert(onConflict = OnConflictStrategy.ABORT)
    fun addMatchUid(uid: String)
}

任何建议表示赞赏。

Answer 1

根据您的上下文，您似乎想要建立one-to-many关系，因此可以遵循以下代码。如果没有，让我知道您到底想要什么。

@Entity(tableName = "user_data")
data class UserData(
    @PrimaryKey(autoGenerate = true) val uid: Int,
    @ColumnInfo(name = "name") var name: String
)

@Entity(tableName = "user_matcher")
data class UserMatcher(
    @PrimaryKey(autoGenerate = true) val uid: Int,
    @ColumnInfo(name = "userId") var userId: Int
)


@Dao
interface UserMatcherDao {
    // Check to see if the uid is in matchedUsers Set/List
    @Query("SELECT m.id as mid, u.id as uid, u.name FROM user_matcher m INNER JOIN user_data u ON m.userId = u.id WHERE m.id = :matchId")
    fun getMatchedUsers(matchId: Int): List<UserData>
}

如何存储可以在房间中查询的字符串？

1 个答案: