我正在尝试实现以下场景 -> 从API获取记录并存储到数据库中。其中一些记录可能会被点击为
favorite
,即使从API获取新数据,它们也应该保留在这个状态。但是,我似乎找不到一种方法来不替换已被选择为favorite
的记录,而不替换整行。我尝试逐个插入记录,并在插入过程中检查记录是否已经存在,但是我似乎无法使用RxJava解决这个问题。
Dao
@Dao
interface KafanaDao {
@Insert
fun insertSingleKafana(kafana: Kafana)
@Insert(onConflict = OnConflictStrategy.REPLACE)
fun insertAll(kafani: List<Kafana>)
@Query("SELECT * FROM ${Constants.KAFANI_TABLE_NAME}")
fun getKafani(): Single<List<Kafana>>
@Query("SELECT * FROM ${Constants.KAFANI_TABLE_NAME} WHERE name = :name")
fun getSingleKafana(name: String): Single<Kafana>
@Query("UPDATE ${Constants.KAFANI_TABLE_NAME} SET isFavorite = :isFavorite WHERE name = :name")
fun setFavourite(name: String, isFavorite: Int)
}
然后我逐一插入记录,但是如何检查它们是否已经存储在数据库中?
fun insertSingleKafanaInDb(kafana: Kafana) {
Observable.fromCallable { kafanaDao.insertSingleKafana(kafana) }
.subscribeOn(Schedulers.io())
.subscribe {
Timber.d("Inserted ${kafana.name} kafani from API in DB...")
}
}
fun getKafaniFromApi(): Observable<List<Kafana>> {
return apiService.getKafani().toObservable().doOnNext {
for (kafana in it) {
//I need to somehow do the check here
insertSingleKafanaInDb(kafana)
}
}
}
实体
@Entity(tableName = Constants.KAFANI_TABLE_NAME)
data class Kafana(
@PrimaryKey
@ColumnInfo(name = "name")
@SerializedName("name")
val name: String,
@ColumnInfo(name = "phone")
@SerializedName("phone")
val phone: String,
@ColumnInfo(name = "address")
@SerializedName("address")
val address: String,
@ColumnInfo(name = "city")
@SerializedName("city")
val city: String,
@ColumnInfo(name = "sponsored")
@SerializedName("sponsored")
val isSponsored: Boolean,
@ColumnInfo(name = "isFavorite")
var isFavorite: Boolean
)
OnConflictStrategy.REPLACE
,它正在被替换。您需要使用使每个条目唯一的方法来检查其是否存在。 - tyczj