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如何对歌曲中的专辑进行排序?

  •  0
  • Vince VD  · 技术社区  · 6 年前

    我试图对专辑ID进行排序,但这并不能解决问题,因为专辑ID显然与对艺术品进行排序无关。

    在这里你可以找到我的代码。

    文斯

     // Columns I'll retrieve from the song table
        String[] columns = {
                SONG_ID,
                SONG_TITLE,
                SONG_ARTIST,
                SONG_ALBUM,
                SONG_ALBUMID,
                SONG_FILEPATH,
        };
    
        // Limits results to only show music files.
        //
        // It's a SQL "WHERE" clause - it becomes `WHERE IS_MUSIC=1`.
        //
        final String musicsOnly = MediaStore.Audio.Media.IS_MUSIC + "=1";
    
        // Querying the system
        cursor = resolver.query(musicUri, columns, musicsOnly, null, null);
    
        if (cursor != null && cursor.moveToFirst())
        {
    
            do {
                // Creating a song from the values on the row
                Song song = new Song(cursor.getInt(cursor.getColumnIndex(SONG_ID)),
                        cursor.getString(cursor.getColumnIndex(SONG_FILEPATH)));
    
                song.setTitle      (cursor.getString(cursor.getColumnIndex(SONG_TITLE)));
                song.setArtist     (cursor.getString(cursor.getColumnIndex(SONG_ARTIST)));
                song.setAlbumID    (cursor.getInt(cursor.getColumnIndexOrThrow(SONG_ALBUMID)));
                song.setAlbum      (cursor.getString(cursor.getColumnIndexOrThrow(SONG_ALBUM)));
                // Using the previously created genre and album maps
                // to fill the current song genre.
                String currentGenreID   = songIdToGenreIdMap.get(Long.toString(song.getId()));
                String currentGenreName = genreIdToGenreNameMap.get(currentGenreID);
                song.setGenre(currentGenreName);
    
                // Adding the song to the global list
                songs.add(song);
            }
            while (cursor.moveToNext());
        }
        else
        {
            // What do I do if I can't find any songs?
    
        }
        cursor.close();
    
    public ArrayList<String> getArtists() {
    
        ArrayList<String> artists = new ArrayList<String>();
    
        for (Song song : songs) {
            String artist = song.getArtist();
    
            if ((artist != null) && (! artists.contains(artist)))
                artists.add(artist);
        }
    
        // Making them alphabetically sorted
        Collections.sort(artists, new Comparator<String>() {
            @Override
            public int compare(String o1, String o2) {
                return o1.compareTo(o2);
            }
        });
        return artists;
    }
    
    /**
     * Returns an alphabetically sorted list with all the
     * albums of the scanned songs.
     *
     * @note This method might take a while depending on how
     *       many songs you have.
     */
    
    public ArrayList<String> getAlbums() {
    
        ArrayList<String> albums = new ArrayList<String>();
    
        for (Song song : songs) {
            String album = song.getAlbum();
    
            if ((album != null) && (! albums.contains(album)))
                albums.add(album);
    
        }
    

    public class Song implements Serializable {
    
    private long id;
    private String data;
    private String title = "";
    private String artist = "";
    private int   albumid      = -1;
    private String album = "";
    private String genre = "";
    
    
    public Song(long songId, String songData){
        this.id = songId;
        this.data = songData;
    }
    
    public long getId(){
        return id;
    }
    public String getData(){return data;}
    
    //Optional meta data
    
    public void setTitle(String title){
        this.title = title;
    }
    public String getTitle() {
        return title;
    }
    
    public void setArtist(String artist){
        this.artist = artist;
    }
    public String getArtist() {
        return artist;
    }
    
    public int getAlbumID() {
        return albumid;
    }
    public void setAlbumID(int albumid) { this.albumid = albumid; }
    
    public void  setAlbum(String album){
        this.album = album;
    }
    public String getAlbum() { return album; }
    
    public void setGenre(String genre) {
        this.genre = genre;
    }
    public String getGenre() {
        return genre;
    }
    
    
    }
    
    5 回复  |  直到 6 年前
        1
  •  1
  •   me_    6 年前

    首先,我不确定为什么在按歌曲存储返回值时要按唱片进行排序(请参见上面的@usman rafi),但是……

    ArrayList<Song> Albums = new Arraylist<>();
    

    不要试图添加流派信息——你不需要它来达到你的目的。

    相册艺术URI可以写为:

    ContentUris.withAppendedId(Uri.parse("content://media/external/audio/albumart"),
        cursor.getInt(cursor.getInt(cursor.getColumnIndexOrThrow(SONG_ALBUMID))));
    

    这将返回唯一的专辑名称(也将只返回专辑中的一首歌曲),如果在MediaStore数据库中重复该专辑(通过拥有同一专辑中的多首歌曲),则只有与查询匹配的第一个实例才会添加到光标中。

    使用排序顺序对相册返回的光标行进行排序;我个人使用SQL的字母顺序(符号、数字、A、B、C…)对其进行排序。

    要编写查询并对其进行排序,我将使用以下代码:

    String[] projection = {"DISTINCT " + MediaStore.Audio.Media.ALBUM_ID,
                             MediaStore.Audio.Media._ID, 
                             MediaStore.Audio.Media.TITLE, 
                             MediaStore.Audio.Media.ARTIST, 
                             MediaStore.Audio.Media.DATA, 
                             MediaStore.Audio.Media.ALBUM,
                             MediaStore.Audio.Media.IS_MUSIC};
    
    String selection = MediaStore.Audio.Media.IS_MUSIC + 
                       "=1 ) GROUP BY (" + MediaStore.Audio.Media.ALBUM_ID;
    
    String sort = MediaStore.Audio.Media.ALBUM + " COLLATE NOCASE ASC";
    
    Cursor cursor = context.
                     getContentResolver().
                     query(MediaStore.Audio.Artists.EXTERNAL_CONTENT_URI,
                     projection,
                     selection,
                     null,
                     sort);
    

    我个人只是循环通过

    if(cursor != null && cursor.getCount >0){
        while(cursor.moveToNext()){
            //create new song item and add the album information you need
            Song album = new Song(cursor.getInt(cursor.getColumnIndex(MediaStore.Audio.Media._ID)),
               cursor.getString(cursor.getColumnIndex(MediaStore.Audio.Media.DATA)));             
    
                album.setAlbumId(cursor.getInt(cursor.getColumnIndex(MediaStore.Audio.Media.ALBUM_ID)));
                album.setAlbumId(cursor.getInt(cursor.getColumnIndex(MediaStore.Audio.Media.ALBUM)));
    
            //add the Song item to the global arraylist
            Albums.add(album)
        }
    }
    
    cursor.close();
    

    Song Album = Albums.get(position);     
    imageView.setURI(ContentUris.withAppendedId(Uri.parse("content://media/external/audio/albumart"),
        Album.getAlbumID());
    

        2
  •  1
  •   Usman Rafi    6 年前

    专辑

    • 可比性
    • 比较器 如果需要实现多个排序逻辑
        3
  •  0
  •   orbbo    6 年前

        4
  •  0
  •   Udit    6 年前

    2、封面和标题不匹配的原因是,您对其中一个进行排序,而忽略了另一个。

    排序前相册-

        5
  •  0
  •   Binary Baba    6 年前

    在代码中不做太多更改的情况下,您可以通过使用比较器对歌曲进行排序来实现您的目的。每当需要使用对象的一个属性(在您的示例中是 QuerySongs

    在你的 getAlbumsId

    public ArrayList<Long> getAlbumsId() {
    
        ArrayList<Long> albumids = new ArrayList<Long>();
        Collections.sort(songs, new Comparator<QuerySongs>() {
            @Override
            public int compare(QuerySongs o1, QuerySongs o2) {
                return o1.getAlbum().compareTo(o2.getAlbum());
            }
        });
    
        for (QuerySongs song : songs){
            Long albumid = song.getAlbumID();
    
            if (! albumids.contains(albumid)) {
                albumids.add(albumid);
            }
        }
    
    
        return albumids;
    }
    

    上面会变异 songs