我正在尝试将地图嵌套到RxJava2中的过滤器中，这样我就可以删除所有没有正确平台的对象。但是不能建造它

Question

override fun getVideoGamesRepository(platformName: String) {
    gameRepo = VideoGameRepositoryImpl()
    compositeDisposable.add(gameRepo.returnVideoGames()
        .filter { gameModel -> gameModel.gamesResults.map {  } }
        .subscribe())
}

下面是我从api中过滤数据的方法。我将只包含相关的数据片段。

我正在尝试将数据库中平台名称的值与我在上一个活动中输入的平台名称进行比较。但在过滤器内部的.map处，.map带红色下划线。

{
"count": 363773,
"next": "https://api.rawg.io/api/games?page=3",
"previous": "https://api.rawg.io/api/games",
"results": [
    {
        "id": 28,
        "slug": "red-dead-redemption-2",
        "name": "Red Dead Redemption 2",
        "released": "2018-10-26",
        "tba": false,
        "background_image": "https://media.rawg.io/media/games/511/5118aff5091cb3efec399c808f8c598f.jpg",
        "rating": 4.54,
        "rating_top": 5,
        "ratings": [
            {
                "id": 5,
                "title": "exceptional",
                "count": 1466,
                "percent": 70.28
            },
            {
                "id": 4,
                "title": "recommended",
                "count": 409,
                "percent": 19.61
            },
            {
                "id": 3,
                "title": "meh",
                "count": 143,
                "percent": 6.86
            },
            {
                "id": 1,
                "title": "skip",
                "count": 68,
                "percent": 3.26
            }
        ],
        "ratings_count": 2056,
        "reviews_text_count": 25,
        "added": 6296,
        "added_by_status": {
            "yet": 284,
            "owned": 3522,
            "beaten": 912,
            "toplay": 951,
            "dropped": 161,
            "playing": 466
        },
        "metacritic": null,
        "playtime": 22,
        "suggestions_count": 630,
        "user_game": null,
        "reviews_count": 2086,
        "saturated_color": "0f0f0f",
        "dominant_color": "0f0f0f",
        "platforms": [
            {
                "platform": {
                    "id": 4,
                    "name": "PC",
                    "slug": "pc",
                    "image": null,
                    "year_end": null,
                    "year_start": null,
                    "games_count": 203916,
                    "image_background": "https://media.rawg.io/media/games/8e0/8e032ac4faf1136e7d708bb3ac61af23.jpg"
                },
                "released_at": "2018-10-26",
                "requirements_en": null,
                "requirements_ru": null
            },
            {
                "platform": {
                    "id": 1,
                    "name": "Xbox One",
                    "slug": "xbox-one",
                    "image": null,
                    "year_end": null,
                    "year_start": null,
                    "games_count": 3085,
                    "image_background": "https://media.rawg.io/media/games/c24/c24ec439abf4a2e92f3429dfa83f7f94.jpg"
                },
                "released_at": "2018-10-26",
                "requirements_en": null,
                "requirements_ru": null
            },
            {
                "platform": {
                    "id": 18,
                    "name": "PlayStation 4",
                    "slug": "playstation4",
                    "image": null,
                    "year_end": null,
                    "year_start": null,
                    "games_count": 4504,
                    "image_background": "https://media.rawg.io/media/games/4be/4be6a6ad0364751a96229c56bf69be59.jpg"
                },
                "released_at": "2018-10-26",
                "requirements_en": null,
                "requirements_ru": null
            }
        ],

Answer 1

filter方法以predicate作为参数。因此您必须为筛选器定义一个返回true或false的条件。

根据你的模型，下面的代码应该返回包含你想要的平台的游戏结果。

compositeDisposable.add(gameRepo.returnVideoGames()
  .map { gameModel ->
    gameModel.gamesResults.filter { gameResult -> gameResult.platforms.any { platform -> platform.name == platformName } }
  }
  .subscribe())

而且，您似乎将整个响应对象存储在您的数据库中。但我认为这是不必要的，也不太实际。相反，您可以存储游戏结果对象。从而让你减少操作次数，达到同样的效果。例如：

compositeDisposable.add(gameRepo.returnVideoGames()
  .filter { gameResult -> gameResult.platforms.any { platform -> platform.name == platformName } }
  .subscribe())

Answer 2

我是个笨蛋，我所需要做的就是以不同的方式查询api调用。很抱歉，伙计们。