所以我的问题是,当我尝试对专辑进行排序时,专辑标题和专辑封面是错误的。
我尝试对专辑ID进行排序,但这并不能解决问题,因为专辑ID显然与艺术排序无关。
当我忽略排序时,一切都是正确的,但是当我尝试对它们进行排序时,专辑名称与专辑封面不匹配。
如何将片段中的专辑排序?
在这里您可以找到我的代码。
提前致谢,
文斯
歌曲模型
// 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); }
歌曲班
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; } }
首先,我不确定在存储按歌曲存储的返回值时为什么要尝试按专辑排序(请参见上面的@Usman Rafi),但是。
将全局数组列表添加到片段的顶部
ArrayList<Song> Albums = new Arraylist<>();
不要尝试添加流派信息-您不需要它来达到目的
专辑封面Uri可以写成:
ContentUris.withAppendedId(Uri.parse("content://media/external/audio/albumart"), cursor.getInt(cursor.getInt(cursor.getColumnIndexOrThrow(SONG_ALBUMID))));
因此,专辑封面和album_id实际上是密不可分的。
所以我的问题是当我尝试对专辑进行排序时…
在查询的选择变量中使用MediaStore.Audio.Media.IS_MUSIC +“ = 1)GROUP BY(” + MediaStore.Audio.Media.ALBUM …
这将返回唯一的专辑名称(也将仅返回专辑中的一首歌曲),如果在您的媒体存储数据库中重复该专辑(通过从同一专辑中获得几首歌曲)重复该专辑,则只会添加与您的查询匹配的第一个实例到您的光标。
对专辑进行排序…
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的字母顺序对其进行排序(符号,数字,a,b,c…。)
您应在此处注意,排序是区分大小写的,除非您指定“ COLLATE NOCASE”
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);
之后,您可以简单地在光标中移动,将每一行添加到所构建的数据对象中,无需进一步排序,并且事物应该以正确的顺序进行。
我个人只是遍历
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();
您现在可以按数组列表中的位置访问排序的专辑信息…您可以使用我在顶部显示的Uri构建器了解专辑封面…
像这样
Song Album = Albums.get(position); imageView.setURI(ContentUris.withAppendedId(Uri.parse("content://media/external/audio/albumart"), Album.getAlbumID());
我希望这对您有用。
我仍然认为您应该建立一个名为Album的数据类