Android房间使用关系搜索和过滤，查询多个表

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我已经为我的pokemonap创建了一个房间数据库，我希望能够根据Pokemon名称和Pokemon类型过滤和搜索数据库

我有一个用于Pokemon实体的表，一个用于PokemonType实体的表和一个用于PokemonTypeJoin实体的连接表，我还有一个数据类PokemonWithTypes，它嵌入了一个Pokemon实体，并定义了它与PokemonType实体列表之间的关系

口袋妖怪实体：

    @TypeConverters(RoomStringListConverter::class)
    @Entity
    data class Pokemon(
        @NotNull
        @PrimaryKey
        @ColumnInfo(name = POKEMON_ID)
        var id: Int,
    
        @ColumnInfo(name = POKEMON_NAME)
        var name: String,
    
        @ColumnInfo(name = POKEMON_URL)
        var url: String,
    
        @ColumnInfo(name = POKEMON_WEIGHT)
        val weight: Int,
    
        @ColumnInfo(name = POKEMON_HEIGHT)
        val height: Int,
    
        @ColumnInfo(name = POKEMON_SPECIES)
        var species: String,
    
        @ColumnInfo(name = POKEMON_MOVES)
        val moves: List<String>
    
    ) 

    const val POKEMON_ID: String = "pokemon_id"
    const val POKEMON_NAME: String = "pokemon_name"
    const val POKEMON_URL: String = "pokemon_url"
    const val POKEMON_HEIGHT: String = "pokemon_height"
    const val POKEMON_WEIGHT: String = "pokemon_weight"
    const val POKEMON_MOVES: String = "pokemon_moves"
    const val POKEMON_SPECIES: String = "pokemon_species"

口袋妖怪加入实体：

    @Entity(primaryKeys = [POKEMON_ID, POKEMON_TYPE_ID])
    class PokemonTypesJoin(
        @NotNull
        @ColumnInfo(name = POKEMON_ID, index = true)
        val pokemon_id: Int,
    
        @NotNull
        @ColumnInfo(name = POKEMON_TYPE_ID, index = true)
        val pokemon_type_id: Int
    
    )
    
    const val POKEMON_ID: String = "id"
    const val POKEMON_TYPE_ID: String = "type_id"

口袋妖怪类

    data class PokemonWithTypes(
        @Embedded
        val pokemon: Pokemon,
        @Relation(
            parentColumn = Pokemon.POKEMON_ID,
            entity = PokemonType::class,
            entityColumn = PokemonType.POKEMON_TYPE_ID,
            associateBy = Junction(
                value = PokemonTypesJoin::class,
                parentColumn = PokemonTypesJoin.POKEMON_ID,
                entityColumn = PokemonTypesJoin.POKEMON_TYPE_ID
            )
        )
        val types: List<PokemonType>
    )

我现在得到了所有的口袋妖怪，可以通过口袋妖怪的名字来搜索，但我希望能够只显示水的类型或草的类型欢迎任何想法

我试着只过滤一个字符串，而不是像这样的查询字符串列表

@Transaction
@Query("SELECT * FROM pokemon, pokemonType WHERE type_name LIKE :filter AND pokemon_name LIKE :search ORDER BY pokemon_id ASC")
fun getPokemonWithTypes(search: String?, filter: String): LiveData<List<PokemonWithTypes>>

@事务
@查询（“从pokemon、pokemonType中选择*，其中键入\u name LIKE:filter和pokemon\u name LIKE:search ORDER BY pokemon\u id ASC”）
有趣的getPokemonWithTypes（搜索：String？，过滤器：String）：LiveData

但它不起作用

---

您可以在这里查看完整内容

我认为@Relation注释不是为该用例设计的。它仅用于返回所有相关类型，而不是经过筛选的子集。我认为你有3个选择：

只需使用Kotlin进行过滤：

pokemonWithTypes.filter{it.types.contains（“GRASS”）}

。我假设你没有超过10000条口袋妖怪的记录，所以性能不是问题

编写一个连接查询。我认为这是为了获得微不足道的性能增益而付出的更多努力

根据使用数据库视图：。这是更静态的，您必须为每种类型编写一个视图

---

谢谢，我将在视图模型中对其进行筛选。他们需要帮助-->

@Transaction
@Query("SELECT * FROM pokemon WHERE pokemon_name LIKE :search ORDER BY pokemon_id ASC")
fun getPokemonWithTypes(search: String?): LiveData<List<PokemonWithTypes>>

PokemonWithTypes(pokemon=Pokemon(id=1, name=bulbasaur, types=[PokemonType(id=4, name=poison, slot=2), PokemonType(id=12, name=grass, slot=1)])
PokemonWithTypes(pokemon=Pokemon(id=4, name=charmander, types=[PokemonType(id=10, name=fire, slot=2), PokemonType(id=12, name=grass, slot=1)])
PokemonWithTypes(pokemon=Pokemon(id=7, name=squirtle, types=[PokemonType(id=11, name=water, slot=2), PokemonType(id=12, name=grass, slot=1)])

@Transaction
@Query("SELECT * FROM pokemon, pokemonType WHERE type_name LIKE :filter AND pokemon_name LIKE :search ORDER BY pokemon_id ASC")
fun getPokemonWithTypes(search: String?, filter: String): LiveData<List<PokemonWithTypes>>