使用Java中的嵌套泛型組織數據（Android）

Question

我正在創建音樂播放器應用程序。我檢索目錄中的數據在一個名爲SongInfo類看起來是這樣的：

public class SongInfo { 

//id of song 
private long _ID; 

//song name 
private String title; 

//artist name 
private String artistName; 

//album name 
private String albumName; 

//song duration 
long duration; 

//year of release 
private String year; 

//constructor 

//comparators to sort data 

我的應用程序具有這樣的結構有一個

ArtistsFragment上顯示點擊藝術家列表 - >它打開具有所選藝術家的專輯列表的專輯片段 - >打開具有所選專輯的歌曲的ArrayList的歌曲片段。我最後做的反而是創建這種類型的數據結構：

//contains most of the info of the song 
ArrayList<SongInfo> songs; 

//contains an arrayList of songs 
ArrayList<ArrayList<SongInfo>> albums; 

//contains an arraylist of albums  
Arraylist<ArrayList<ArrayList<SongInfo>>> artists; 

這我非常肯定是錯誤的方式做到這一點，因爲如果我有什麼更多的層？我正在尋找更好的方法來做到這一點。我只需要一個提示，但是插圖/解釋會很棒。謝謝:)

Answer 1

假設我們要實現這些類：

public static class Song 
{ 
    String title; 
    Artist artist; 
    Album album; 
    Song(String title, Artist artist, Album album) 
     { this.title = title; this.artist = artist; this.album = album; } 
    @Override public String toString() { return "Song: " + title; } 
} 

public static class Artist 
{ 
    String name; 
    Artist(String name) { this.name = name; } 
    @Override public String toString() { return "Artist: " + name; } 
} 

public static class Album 
{ 
    String name; 
    Album(String name) { this.name = name; } 
    @Override public String toString() { return "Album: " + name; } 
} 

其中：

• Song就是你現在用相當不幸的名稱SongInfo
• Artist調用是一個類，你還沒有，但你應該。如果你不能處理這個問題，那麼只要你看到Artist就想到一個String，裏面有一個藝術家的名字。
• Album是另一類，你還沒有，但你應該。如果您無法處理該問題，那麼無論何時您看到Album只需考慮一個String並附上相冊的標題即可。

並假設我們開始與這個數據集：

Artist artist1 = new Artist("Artist1"); 
    Artist artist2 = new Artist("Artist2"); 
    Set<Artist> artists = new HashSet<>(Arrays.asList(artist1, artist2)); 
    Album album1 = new Album("Album1"); 
    Album album2 = new Album("Album2"); 
    Set<Album> albums = new HashSet<>(Arrays.asList(album1, album2)); 
    Song song11a = new Song("Song11a", artist1, album1); 
    Song song11b = new Song("Song11b", artist1, album1); 
    Song song22a = new Song("Song22a", artist2, album2); 
    Song song22b = new Song("Song22b", artist2, album2); 
    List<Song> songs = Arrays.asList(song11a, song11b, song22a, song22b); 

那麼下面會給你想要的東西：

Collection<Song> getSongsByArtist(Collection<Song> songs, Artist artist) 
{ 
    return songs.stream().filter(song -> song.artist == artist) 
     .collect(Collectors.toList()); 
} 

Collection<Album> getAlbumsByArtist(Collection<Song> songs, Artist artist) 
{ 
    return songs.stream().filter(song -> song.artist == artist) 
     .map(song -> song.album).collect(Collectors.toSet()); 
} 

和下面會給你你一直想知道的一切關於您的數據集，但不敢問：

Map<Song,Artist> artistsBySong = songs.stream() 
    .collect(Collectors.toMap(Function.identity(), song -> song.artist)); 
Map<Song,Album> albumsBySong = songs.stream() 
    .collect(Collectors.toMap(Function.identity(), song -> song.album)); 
Map<Artist,List<Song>> songsByArtist = songs.stream() 
    .collect(Collectors.groupingBy(song -> song.artist)); 
Map<Album,List<Song>> songsByAlbum = songs.stream() 
    .collect(Collectors.groupingBy(song -> song.album)); 
assert songs.stream().map(song -> song.album) 
    .collect(Collectors.toSet()).equals(albums); 
assert songsByAlbum.keySet().equals(albums);